Revision: Trigonometry >> Trigonometrical Identities Maths (English Medium) ICSE Class 10 CISCE

When an equation, involving trigonometrical ratios of an angle A, is true for all values of A, the equation is called a trigonometric identity. 

\[sineA=\frac{\text{Perpendicular}}{\text{Hypotenuse}}\]

\[cosineA=\frac{\mathrm{Base}}{\text{Hypotenuse}}\]

\[tangentA=\frac{\text{Perpendicular}}{\mathrm{Base}}\]

\[cotangent A = \frac{\text{Base}}{\text{Perpendicular}}\]

\[secantA=\frac{\text{Hypotenuse}}{\mathrm{Base}}\]

\[cosecantA=\frac{\text{Hypotenuse}}{\text{Perpendicular}}\]

\[\sin\mathrm{A}=\frac{1}{\mathrm{cosec~A}}\quad\mathrm{and}\quad\mathrm{cosec~A}=\frac{1}{\sin\mathrm{A}}\]

\[\cos\mathrm{A}=\frac{1}{\sec\mathrm{A}}\quad\mathrm{and}\quad\mathrm{sec}\mathrm{A}=\frac{1}{\cos\mathrm{A}}\]

\[\tan\mathrm{A}=\frac{1}{\cot\mathrm{A}}\quad\mathrm{and}\quad\cot\mathrm{A}=\frac{1}{\tan\mathrm{A}}\]

• sin⁡θ⋅cosec⁡θ = 1

• cos⁡θ⋅sec⁡θ = 1

• tan⁡θ⋅cot⁡θ = 1

\[tanA=\frac{\sin A}{\cos A}\]

\[cotA=\frac{\cos A}{\sin A}\]

For an acute angle A, 

1. sin (90° - A) = cos A
2. cos (90° - A) = sin A
3. tan (90° - A) = cot A
4. cot (90° - A) = tan A
5. sec (90° - A) = cosec A
6. cosec (90° - A) = sec A

`cot theta = "𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒"/"𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒"`

Let x be the hypotenuse by applying Pythagoras theorem.

𝐴𝐶² = 𝐴𝐵² + 𝐵𝐶²

𝑥² = 16 + 9

`x^2 = 25 => x = 5`

`sec theta = (AC)/(BC) = 5/3`

`cosec theta = (AC)/(AB) = 5/4`

On substituting in equation we get

`sqrt((sec theta - cosec theta)/(sec theta + cosec theta)) = sqrt((5/3 - 5/4)/(5/3 + 5/4))`

`= sqrt(((20 - 15)/12)/((20 + 15)/12)) = sqrt(5/35) = 1/sqrt7`

We have `sin theta = 3/4`

In ΔABC

`AC^2 = AB^2 + BC^2`

`=> (4)^2 = (3)^2 + BC^2`

`=> BC^2= 16 - 9`

`=> BC^2 = 7`

`=> BC = sqrt7`

`:. cosec theta = 4/3, sec theta = 4/sqrt7 and cot theta = sqrt7/3`

Now

L.H.S `sqrt((cosec^2 theta - cot^2 theta)/(sec^2 theta - 1))`

`= sqrt(((4/3)^2 - (sqrt7/3)^2)/((4/sqrt7)^2 - 1)`

`= sqrt((16/9 - 7/9)/(16/7 - 1)`

`=sqrt((9/9)/((16 - 7)/7 ))`

`= sqrt(7/9)`

`= sqrt7/3`

= R.H.S

Let `(a sin  theta - b cos theta)/(a sin theta + b cos theta)`

Divide both Nr and Dr with cos θ of (a)

`((a sin theta - b cos theta)/cos theta)/((a sin theta + b cos theta)/cos theta)`

`= (tan theta - b)/(a tan theta + b)`

`=(a xx (a/b) - b)/(a xx (a/b) + b)`

`= (a^2 - b^2)/(a^2 + b^2)`

Given `cos theta + cos^2 theta = 1`

We have to prove sin¹² θ + 3 sin¹⁰ θ + 3 sin⁸ θ + sin⁶ θ + 2 sin⁴ θ + 2 sin² θ − 2 = 1

From the given equation, we have

`cos theta + cos^2 theta = 1`

`=> cos theta = 1 - cos^2 theta`

`=> cos theta = sin^2 theta`

`=> sin^2 theta = cos theta`

Therefore, we have

sin¹² θ + 3 sin¹⁰ θ + 3 sin⁸ θ + sin⁶ θ + 2 sin⁴ θ + 2 sin² θ − 2

`= (sin^12 theta + 3 sin^10 theta + 3 sin^8 theta + sin^6 theta) + (2 sin^4 theta + 2 sin^2 theta) - 2` 

`= {(sin^4 theta)^3 + 3(sin^4 theta)^2 sin^2 theta + 3 sin^4 theta(sin^2 theta)^2 + (sin^2 theta)^3} + 2(sin^4 theta + sin^2 theta) - 2`

`= (sin^4 theta  + sin^2 theta)^3 + 2 (sin^4 theta + sin^2 theta) - 2`

`= (cos^2 theta + cos theta)^3 + 2 (cos^2 theta + cos theta) - 2`

`= (1)^3 + 2(1) - 2`

= 1

Hence proved

We need to prove `(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`

Using the property `1 + tan^2 theta = sec^2 theta` we get

`(1 + tan^2 A)+(1 + 1/tan^2 A) = sec^2 A = ((tan^2 A + 1)/tan^2 A)`

`= sec^2 A + (sec^2 A)/(tan^2 A)`

Now using `sec theta = 1/cos theta` and `tan theta = sin theta/cos theta` we get

`sec^2 A + ((sec^2 A)/(tan^2 A)) = 1/cos^2 A + ((1/cos^2 A)/((sin^2 A)/(cos^2 A)))`

`= 1/cos^2 A + (1/cos^2A  xx cos^2 A/sin^2 A)` 

` = 1/cos^2 A + 1/sin^2 A`

`= (sin^2 A + cos^2 A)/(cos^2 A(sin^2 A))`

Further, using the property, `sin^2 theta + cos^2 theta = 1` we get

`(sin^2 A + cos^2 A)/(cos^2 A(sin^2 A)) = 1/(cos^2 A (sin^2 A))`

`= 1/((1 - sin^2 A)(sin^2 A))`     (using `cos^2 theta = 1 - sin^2 theta`)

`= 1/(sin^2 A - sin^4 A)`

Hence proved

L.H.S = `cot^2 A cosec^2B - cot^2 B cosec^2 A`

`= cot^2 A(1+ cot^2 B) - cot^2 B(1 + cot^2 A)`    (∵ `1 + cot^2 theta = cosec^2 theta`)

`= cot^2 A + cot^2 A cot^2 B - cot^2 B - cot^2 A cot^2 B`

`= cot^2 A - cot^2 B`

Hence proved

`(tan^3 theta)/(1 + tan^2 theta) + (cot^3 theta)/(1 + cot^2 theta) `         [`∵ sec^2 theta - tan^2 theta = 1 - cosec^2 theta - cot^2 theta = 1`]

`= tan theta + cos^2 theta = cot^3 theta xx sin^3 theta`

`[∵ 1/sec^2 theta = cos^2 theta, 1/cosec^2 theta = 1 + cot^2 theta]`

`sin^3 theta/cos^3 theta xx cos^2 theta + cos^3 theta/sin^3 theta xx sin^2 theta`

`sin^3 theta/cos theta + cos^3 theta/sin theta`

`= (sin^4 theta + cos^4 theta)/(sin theta cos theta)`

` (1 - 2sin^2 theta cos^2 theta)/(sin theta cos theta)`

`1/(sin theta cos theta) - (2 sin^2 theta cos^2 theta)/(sin theta cos theta)`

`sec theta cosec theta - 2sin theta cos theta`.

We have to prove `tan A/(1 + tan^2  A)^2 + cot A/((1 + cot^2 A)) = sin A  cos A`

We know that `sin^2 A + cos^2 A = 1`

So

`tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2`

`= tan A/(sec^2 A)^2  + cot A/(cosec^2 A)^2`

`= tan A/sec^4 A  + cot A/(cosec^4 A)`

`= (sin A/cos A)/(1/cos^4 A) + (cos A/sin A)/(1/sin^4 A)`

`= (sin A cos^4 A)/cos A + (cos A sin^4 A)/sin A`

`= sin A cos^3 A + cos A sin^3 A`

`= sin A cos A (cos^2 A + sin^2 A)`

= sin A cos A

Hence proved.

`= (a cos^3 theta + 3a cos theta sin^2 theta + a sin^3 theta + 3a cos^2 theta sin theta)^(3/2) + (a cos^3 theta + 3a cos theta sin^2 theta - a sin^3 theta - 3a cos^2 theta sin theta)^(2/3)`

`= a^(1/3) (cos^3 theta + 3 cos theta sin^2 theta + sin^3 theta + 3 cos^2 theta sin theta)^(2/3) + a^(2/3) (cos^3 theta + 3 cos theta sin^2 theta + sin^3 theta - 3 cos^2 theta sin theta)^(2/3)`

`= a^(1/3) [(cos theta + sin theta)^3]^(2/3) + a^(2/3) (cos theta - sin theta)^3]^(2/3)`

`= a^(2/3) [(cos theta + sin theta)^2] + a^(2/3) (cos theta - sin theta)^2`

`= a^(2/3) [cos^2 theta + sin^2 theta - 2sin theta cos theta]`

`= a^(2/3) [cos^2 theta + sin^2 theta + 2 sin theta cos theta] +_ a^(2/3) [cos^2 theta + sin^2 theta - 2 sin theta cos theta]`

`= a^(2/3) [1 + 2 sin theta cos theta] + a^(2/3)[1 - 2 sin theta cos theta]`

`= a^(2/3) [1 + 2 sin theta cos theta + 1  - 2 sin theta cos theta]`

`= a^(1/3) (1 + 1) = 2a^(2/3)`

R.H.S

L.H.S. = `(cos^2θ)/(sinθ) + sin θ` 

= `(cos^2θ + sin^2θ)/(sin θ)`

= `1/(sin θ)`   ...[∵ sin²θ + cos²θ = 1]

= cosec θ

= R.H.S.

∴ `(cos^2θ)/(sin θ) + sin θ = "cosec"  θ`

We have to prove sec A(1 − sin A)(sec A + tan A) = 1

We know that sec² A − tan² A − 1 

So,

sec A(1 − sin A)(sec A + tan A) = {sec A(1 − sin A)}(sec A + tan A)

= (sec A − sec A sin A)(sec A + tan A)

= `(sec A - 1/cos A  sin A) (sec A + tan A)`     ...`(∵ sec theta = 1/costheta)`

= `(sec A - sin A/cos A) (sec A + tan A)`    ...`(∵ tan theta = sin theta/costheta)`

= (sec A − tan A)(sec A + tan A)

= sec² A − tan² A

= 1 = R.H.S.    ... (∵ sec² θ = 1 tan² θ)

L.H.S. = sec²θ + cosec²θ

= `1/(cos^2θ) + 1/(sin^2θ)`

= `(sin^2θ + cos^2θ)/(cos^2θ.sin^2θ)`

= `1/(cos^2θ.sin^2θ)`   ...[∵ sin²θ + cos²θ = 1]

= `1/(cos^2θ) xx 1/(sin^2θ)`

= sec²θ × cosec²θ

= R.H.S.

∴ sec²θ + cosec²θ = sec²θ × cosec²θ

L.H.S. = `secA/(secA + 1) + secA/(secA - 1)`

= `(sec^2A - secA + sec^2A + secA)/(sec^2A - 1`

= `(2sec^2A)/tan^2A`   ...(∵ sec² A – 1 = tan² A)

= `(2/cos^2A)/(sin^2A/cos^2A)`

= `2/sin^2A`

= 2 cosec² A = R.H.S.

LHS= `(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos ^3 θ - sin^3 θ)/(cos θ - sin θ) `

=` ((cos θ + sin θ)(cos^2 θ - cos θ sin θ + sin^2 θ))/((cos θ + sin θ)) + ((cos θ - sin θ)(cos^2 θ + cos θ sin θ + sin^2 θ))/((cos θ - sin θ))`

= (cos² θ + sin² θ − cos θ sin θ) + (cos² θ + sin² θ + cos θ sin θ)`

= (1 − cos θ sin θ) + (1 + cos θ sin θ)

= 2

= RHS

Hence, LHS = RHS

L.H.S. = `(sin θ + tan θ)/(cos θ)`

= `(sin θ)/(cos θ) + (tan θ)/(cos θ)`

= tan θ + tan θ sec θ

= tan θ (1 + sec θ)

= R.H.S.

∴ `(sin θ + tan θ)/(cos θ) = tan θ (1 + sec θ)`

L.H.S. = `(cosθ)/(1 + sinθ)`

= `(cosθ)/(1 + sinθ) xx (1 - sinθ)/(1 - sinθ)`   ...[On rationalising the denominator]

= `(cosθ(1 - sinθ))/(1 - sin^2θ)`

= `(cosθ(1 - sinθ))/(cos^2θ)`   ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 -sin^2θ = cos^2θ)]`

= `(1 - sinθ)/(cosθ)`

= R.H.S.

∴ `(cosθ)/(1 + sinθ) = (1 - sinθ)/(cosθ)`

We know that `sin^2 theta + cos^2 theta = 1`

So,

LHS = `"cosec" theta sqrt(1 - cos^2 theta)`

= `"cosec" theta sqrt (sin^2 theta)`

= cosec θ . sin θ

`1/sin theta xx sin theta`

= 1

= RHS hence proved.

LHS = `1 + (tan^2 θ)/((1 + sec θ))`

=` 1 + ((sec^2 θ - 1))/((sec theta + 1))`

=`1 + ((sec theta + 1)(sec theta - 1))/((sec theta + 1))`

=`1 + (sec theta - 1)`

= sec θ

LHS = RHS

LHS =`1/((1+ sin θ)) + 1/((1 - sin θ))`

= `((1 - sin θ) + (1 + sin θ))/((1 + sin θ)(1 - sin θ))`

= `2/(1 - sin^2 θ)`

= `2/(cos^2 θ)`

= 2 sec² θ

= RHS

Hence Proved.

L.H.S. = cot θ + tan θ

L.H.S. = `costheta/sintheta + sintheta/costheta`

L.H.S. = `(cos^2theta + sin^2theta)/(sintheta costheta)`

[cos² θ + sin² θ = 1]

L.H.S. = `1/(sintheta costheta)`

Use Reciprocal Identities:

The expression can be split into `(1/sin θ) xx (1/cos θ)`.

`1/sin θ` = cosec θ

`1/cos θ` = sec θ

L.H.S. = cosec θ.sec θ

L.H.S. = sec θ.cosec θ

∴ L.H.S. = R.H.S.

LHS = `sqrt((1 + sin θ)/(1 - sin θ))`

=`sqrt(((1 + sin θ))/(1 - sin θ) xx ((1 + sin θ))/(1 + sin θ))`

=` sqrt(((1 + sin θ)^2)/(1 - sin^2 θ))`

=`sqrt(((1 + sin θ)^2)/(cos^2 θ))`

=`(1 + sin θ)/cos θ`

=`1/cos θ + (sin θ)/(cos θ)`

= sec θ + tan θ

= RHS

`"LHS" = sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ))`

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get,

`"LHS" = sqrt((1 + sin θ)/(1 - sin θ) × (1 + sin θ)/(1 + sin θ)) + sqrt((1 - sin θ)/(1 + sin θ) × (1 - sin θ)/(1 - sin θ))`

`"LHS" = sqrt((1 + sin θ)^2/(1 - sin^2 θ)) + sqrt((1 - sin θ)^2/(1 - sin^2 θ))`

`"LHS" = sqrt((1 + sin^2θ)/(1 - sin^2 θ)) + sqrt((1 - sin^2θ)/(1 - sin^2 θ))`

`"LHS" = sqrt((1 + sin^2θ)/(cos^2 θ)) + sqrt((1 - sin^2θ)/(cos^2 θ))`

`"LHS" = (1 + sin θ)/(cos θ) + (1 - sin θ)/(cos θ)`

`"LHS" = (1 + cancel(sin θ) + 1 -cancel(sin θ))/(cos θ)`

LHS = `2/(cos θ)`

LHS = 2. `1/(cos θ)`

LHS = 2. sec θ

RHS = 2. sec θ

LHS = RHS

Hence proved.

LHS = `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ)`

= `(sin θ(1 - 2sin^2 θ))/(cos θ(2 cos^2 θ - 1))`

= `(tan θ(1 - 2(1 - cos^2 θ)))/(2 cos^2θ - 1 )`

= `(tan θ(1 - 2 + 2 cos^2 θ))/(2 cos^2θ - 1 )`

= `(tan θ(2 cos^2 θ - 1))/(2 cos^2θ - 1 )`

= tan θ

= RHS

Hence proved.

LHS = tan² θ − sin² θ

= `sin^2 θ/cos^2 θ - sin^2 θ`   `[∵ tan^2 θ = sin^2 θ/cos^2 θ]`

`=> sin^2 θ [1/cos^2 θ- 1]`

`sin^2 θ [(1 - cos^2 θ)/cos^2 θ]`

`=> sin^2 θ. sin^2 θ/cos^2 θ = sin^2 θ tan^2 θ `

LHS = RHS

Hence proved

L.H.S. = `sqrt(((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ)))`

= `sqrt((1 + sin^2θ - 2sinθ)/(1 - sin^2θ)`

= `sqrt((1 + sin^2θ - 2sinθ)/(cos^2θ)`

= `sqrt( 1/cos^2θ + sin^2θ/cos^2θ - (2sin θ)/cos θ xx 1/cosθ`

= `sqrt( sec^2θ + tan^2 θ - 2 tan θ . sec θ)`

= `sqrt((sec θ - tan θ)^2)`

= sec θ – tan θ

= R.H.S.

Hence proved.

L.H.S. = `sqrt((1 - sin θ)/(1 + sin θ))`

= `sqrt(((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ))`

= `sqrt(((1 - sin θ)^2)/(1 - sin^2θ)`

= `sqrt(((1 - sin θ)^2)/(cos^2θ)`

= `(1 - sin θ)/(cos θ)`

= `1/(cos θ) - (sin θ)/(cos θ)`

= sec θ – tan θ

= R.H.S.

Hence Proved.

LHS = `(cos theta   cosec theta - sin theta  sec theta )/(costheta + sin theta)`

      =`((cos theta   sin theta)/(sin theta cos theta))/(cos theta + sin theta)`

      =`(cos^2  theta - sin^2 theta)/(cos theta  sin theta ( cos theta + sin theta))`

      =`((cos theta + sin theta )( cos theta - sin theta))/(cos theta sin theta ( cos theta + sin theta))`

      =`((cos theta - sin theta ))/(cos theta   sin theta)`

     =`1/ sin theta - 1/ cos theta`

     =`cosec theta - sec theta`

     = RHS
Hence, LHS = RHS 

L.H.S. = `tanA/(1 - cotA) + cotA/(1 - tanA)`

= `tanA/(1 - 1/tanA) + (1/tanA)/(1 - tanA)`

= `tan^2A/(tanA - 1) + 1/(tanA(1 - tanA))`

= `(tan^3A - 1)/(tanA(1 - tanA))`

= `((tanA - 1)(tan^2A + 1 + tanA))/(tanA(tanA - 1)`

= `(sec^2A + tanA)/tanA`

= `(1/cos^2A)/(sinA/cosA) + 1`

= `1/(sinAcosA) + 1`

= sec A cosec A + 1 = R.H.S.

We have `(cot theta + tan theta ) = m and ( sec theta - cos theta )=n`

Now, `m^2 n = [(cot theta + tan theta )^2 (sec theta -  cos theta )]`

                  =`[(1/tan theta + tan theta )^2 (1/cos theta- cos theta )]`

                  =`(1+tan^2 theta)^2/tan^2 theta xx ((1-cos^2 theta))/costheta`

                  =`sec^4 theta/tan^2 theta xx sin^2 theta/ cos theta`

                  =`sec ^4 theta /(sin^2 theta/cos^2 theta) xx sin^2 theta / cos theta`

                  =`(cos^2 xxsec^4 theta)/costheta`

                  =`cos theta sec^4 theta`

                 =`1/ sec theta xx sec ^4 theta = sec^3 theta`

∴`(m^2 n)^(2/3) =(sec^3 theta )^(2/3) =  sec^2 theta`

Again, `mn^2 = [(cot theta + tan theta )( sec theta - cos theta )^2 ]`

                      =`[(1/tan theta + tan theta).(1/ cos theta - cos theta)^2]`

                     =`((1+ tan^2 theta))/tan theta xx ((1- cos^2 theta)^2)/cos^2 theta `

                     =`sec^2 theta/tan theta xx sin^4 theta/cos^2 theta`

                    =`sec^2 theta/(sintheta/costheta) xx sin^4 theta/ cos^2 theta`

                    =`(sec^2 xx sin^3 theta)/cos theta`

                     =`1/ cos^2 theta xx sec^3 theta/ cos theta = tan^3 theta `

∴ `(mn^2)^(2/3) = (tan ^3 theta )^(2/3) = tan^2 theta`

Now,` (m^2n)^(2/3) - (mn^2)^(2/3)`

                   =`sec^2 theta - tan^2 theta =1 `

                   =RHS

Hence proved.           

We have `( cosec theta - sin theta ) = a^3`

      = > ` a^3 = (1/ sin theta - sin theta)`

      = > `a^3 = ((1- sin^2 theta))/sin theta = cos^2 theta / sin theta`

∴ `a=(cos^(2/3) theta)/(sin ^(1/3) theta)`

Again, `(sec theta - cos theta ) = b^3`

       = >`b^3 = (1/cos theta - cos theta )`

      =` ((1-cos^2 theta))/ cos theta`

      =` (sin^2 theta)/cos theta`

∴ b =` (sin ^(2/3) theta)/(cos ^(1/3) theta)`

Now , LHS  = `a^2 b^2 (a^2 + b^2 ) `

  =` a^3 (ab^2) + ( a^2 b^2 ) b^3 `

=`a^3 ( ab^2 ) + ( a^2 b^2 ) b^3 `

=`(cos^2 theta)/(sin theta) xx [(cos ^(2/3) theta)/(sin^(1/3) theta) xx (sin ^(4/3)theta)/(cos ^(2/3) theta)] + [ ( cos ^(4/3) theta theta)/(sin ^(2/3) theta)xx(sin^(2/3)theta)/(cos ^(1/3)theta)] xx sin^2 theta/ cos theta`

 =`cos^2 theta / sin theta xx sin theta + cos theta xx sin^2theta / costheta`

 =`cos^2 theta + sin^2 theta = 1`

= RHS
Hence, proved

LHS = cosec⁴ θ − cosec² θ

LHS = cosec² θ (cosec² θ − 1)

LHS = (cot² θ + 1)cot² θ     ...`{(cot^2 θ + 1 = cosec^2 θ),(∵ cot^2 θ = cosec^2 θ - 1):}`

LHS = cot⁴ θ + cot² θ

RHS = cot⁴ θ + cot² θ

RHS = LHS 

Hence proved.

RHS = cot⁴ θ + cot² θ

RHS = cot² θ (cot² θ + 1) 

RHS = (cosec² θ − 1)cosec² θ  ...`{(cot^2 θ + 1=cosec^2 θ),(∵ cot^2θ = cosec^2 θ - 1):}`

RHS = cosec⁴ θ − cosec² θ

LHS = cosec⁴ θ − cosec² θ

RHS = LHS 

Hence proved.

LHS = `(1 + tan theta + cot theta )(sintheta - cos theta) `

= `sin theta + tan theta sin theta + cot theta sin theta - cos theta - tan theta cos theta - cot theta cos theta `

= `sin theta + tan theta sin theta + cos theta/sin theta xx sin theta - cos theta - sin theta/cos theta xx cos theta - cot theta cos theta`

= `sin theta + tan theta  sin theta + cos theta - cos theta - sin theta - cot theta cos theta`

= `tan theta sin theta - cot theta cos theta`

= `sin theta/cos theta xx 1/(cosec theta) - cos theta/sin theta xx 1/sec theta`

= `1/(cosec theta) xx 1/(cosec theta) xx sec theta - 1/sec theta xx 1/sec theta xx cosec theta`

= `sec theta/(cosec^2 theta) - (cosec theta)/sec^2 theta`

= RHS

Hence, LHS = RHS

We have to prove `sin^2 A cot^2 A + cos^2 A tan^2 A = 1`

We know that `sin^2 A + cos^2 A = 1`

So,

`sin^2 A cot^2 A  + cos^2 A tan^2 A = sin^2 A (cos^2 A)/(sin^2 A) + cos^2  A(sin^2 A)/(cos^2 A)`

`= cos^2 A + sin^2 A`

= 1

We have `(tan theta + sin theta ) = m and ( tan theta - sin theta )=n`

 Now,LHS = `(m^2-n^2)^2`

                 =`[(tan^2 theta + sin theta )^2 - "( tan theta - sin theta )^2]^2`

                =`[(tan^2 theta + sin^2 theta + 2 tan  theta  sin theta )-( tan^2 theta + sin^2 theta -2 tan theta sin theta )]^2`

               =`[(tan^2 theta +sin^2 theta + 2 tan theta sin theta - tan^2 theta -  sin^2 theta+ 2 tan theta sin theta )]^2`

              =`(4 tan theta sin theta )^2`

              =`16 tan^2 theta sin^2 theta`

              =`16 (sin ^2 theta )/(cos^2 theta ) sin^2 theta`

              =`16 ((1- cos^2 theta) sin ^2 theta)/ cos^2 theta`

               =` 16 [ tan^2 theta (1- cos^2 theta)]`

               =`16 (tan^2 theta - tan^2 theta cos^2 theta)`

               =`16 (tan^2 theta -(sin^2 theta)/(cos^2 theta) xx cos^2 theta )s`

              =`16 ( tan^2 theta - sin^2 theta )`

              =`16 (tan theta + sin theta ) ( tan theta - sin theta)`

              =`16 mn                 [(tan theta + sin^theta )( tan theta - sin theta ) =mn]`

               =`∴ (m^2 - n^2 )(m^2 - n^2 )^2 = 16 mn`          

We have `x^2 - y^2 = [( a sec theta + b tan theta )^2 - ( a tan  theta + b sec theta )^2]`

= `(a^2 sec^2 theta + b^2 tan^2 theta + 2 ab sec theta tan theta) - (a^2 tan^2 theta + b^2 sec^2 theta + 2 ab tan theta sec theta)`

= `a^2 sec^2 theta + b^2 tan^2 theta - a^2 tan^2 theta - b^2 sec^2 theta`

= `(a^2 sec^2 theta - a^2 tan^2 theta)-( b^2 sec^2 theta - b^2 tan ^2 theta)`

= `a^2 ( sec^2 theta - tan^2 theta )-b^2 ( sec^2 theta - tan^2 theta)`

= `a^2 - b^2  [∵ sec^2 theta - tan^2 theta =1]`

Hence, `x^2 - y^2 = a^2 - b^2`

LHS =` (1+ cos theta + sin theta)/(1+ cos theta-sin theta)`

       =` ({(1+cos theta)+ sin theta}{(1+ cos theta)+ sin theta})/({(1+ cos theta )-sin theta}{(1+ cos theta )+ sin theta}) {"Multiplying the numerator and denominator by "(1 + costheta +sin theta}`

     =`({(1+ cos theta)+ sin theta}^2)/({(1+ cos theta )^2-sin ^2 theta})`

     =`(1+ cos^2 theta + 2 cos theta + sin ^2 theta + 2 sin theta (1+ cos theta))/(1+ cos^2 theta + 2 cos theta - sin ^2 theta)`

     =`(2+2 cos theta + 2 sin theta (1+ cos theta))/(1+ cos ^2 theta + 2 cos theta -(1-cos^2 theta))`

     =`(2(1+ cos theta)+2sin theta (1+ cos theta))/(2 cos^2 theta+2 cos theta)`

      =`(2(1+ cos theta) (1+ sin theta))/( 2 cos theta (1+ cos theta))`

      =`(1+sin theta)/cos theta`

      = RHS

Here, `(sec theta + tan theta ) /( sec theta - tan theta)`

      =`((sec theta + tan theta ) ( sec theta + tan theta))/(( sec theta - tan theta ) ( sec theta + tan theta ))`

      =` ((sec theta + tan theta )^2) /( sec^2 theta - tan^2 theta)`

      =`((sec theta + tan theta )^2)/1`

      =`(sec theta + tan theta )^2`

 Again , `(sec theta + tan theta )2`

      =` sec^2 theta + tan^2 theta + 2 sec theta  tan theta `

      =` 1+ tan^2 theta + tan^2 theta + 2 sec theta tan theta`

      =`1+2 tan^2 theta + 2 sec theta tan theta `

We have `m^2 + n^2 = [(a cos theta + b sin theta)^2 + (a sin theta - b cos theta)^2]`

= `(a^2 cos^2 theta + b^2 sin ^2 theta + 2 ab cos theta sin theta) + (a^2 sin^2 theta + b^2 cos^2 theta -2ab cos theta sin theta)`

= `a^2 cos^2 theta + b^2 sin^2 theta + a^2 sin^2 theta + b^2 cos^2 theta`

= `(a^2 cos^2 theta + b^2 sin^2 theta) + (b^2 cos^2 theta + b^2 sin^2 theta)`

= `a^2 (cos^2 theta + sin^2 theta ) + b^2 (cos^2 theta + sin^2 theta)`

= `a^2 + b^2    [∵ sin^2 + cos^2 = 1]`

Hence, `m^2 + n^2 = a^2 + b^2`

LHS = `(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))`

= `(cos^2 theta/sin^2 theta(1/costheta-1))/((+ sin theta)) + (1/cos^2 theta(sin theta -1))/((1+ 1/cos theta))`

= `((cos^2 theta)/(sin^2 theta )((1- cos theta)/(cos theta)))/((1+sin theta))+ (((sin theta -1 ))/(cos ^2theta ))/(((cos theta + 1 )/(cos theta)))`

= `(cos^2 theta (1- cos theta))/(sin^2 theta cos theta (1+ sin theta))+ ((sin theta -1) cos theta)/((cos theta +1 ) cos^2 theta)`

= `(cos theta (1-cos theta))/((1- cos^2 theta)(1+ sin theta)) + ((sin theta -1)cos theta)/((costheta + 1 ) (1- sin^2 theta))` 

= `(cos theta (1-cos theta))/((1- cos theta )( 1+ cos theta )(1+ sin theta)) + (-(1 sin theta ) cos theta)/((cos theta +1)(1-sin theta )(1+ sin theta))`

 = `cos theta/((1+ cos theta )(1+ sin theta)) - cos theta/((1 + cos theta)(1+ sin theta))`

= 𝜃
= RHS

Given : `cos A + cos^2 A = 1`

we have to prove `sin^2 A + sin^4 A = 1`

Now

`cos A + cos^2 A = 1`

`=>cos A = 1 - cos^2 A`

`=> cos A = sin^2 A`

`=> sin^2 A = cos A`

Therefore, we have

`sin^2 A + sin^4 A = cos A + (cos A)^2`

`= cos A + cos^2 A`

= 1

Hence proved.

We have `m^2 + n^2 = [(a cos theta + b sin theta)^2 + (a sin theta - b cos theta)^2]` 

`= (a^2 cos^2 theta + b^2 sin^2 theta + 2ab cos theta sin theta) + (a^2sin^2 theta + b^2cos^2 theta - 2ab sin theta cos theta)`

`= a^2 cos^2 theta + b^2 sin^2 theta + a^2 sin^2 theta + b^2 cos^2 theta`

`= (a^2 cos^2 theta + a^2 sin^2 theta) + (b^2 cos^2 theta + b^2 sin^2 theta)`

`= a^2(cos^2 theta + sin^2 theta) + b^2(cos^2 theta + sin^2 theta)`

`=a^2 + b^2`   [∵ `sin^2 theta + cos^2 theta = 1]`

Hence, m² + n² = a² + b²

LHS= `(sin theta+1cos theta)/(cos theta-1+sin theta) `

      =`((sin theta+1-cos theta)(sin theta+cos theta+1))/((cos theta -1 + sin theta)(sin theta + cos theta +1))`

      =`((sin theta + 1 )^2 - cos^2 theta)/((sin theta + cos theta )^2 -1^2)`

       =`(sin^2 theta +1+2 sin theta - cos^2 theta)/(sin^2 + cos^2 theta+2 sin theta  cos theta -1)`

      =`(sin^2 theta + sin^2 theta + cos^2 theta +2sin theta - cos^2 theta)/(2 sin theta   cos theta)`

      =`(2 sin ^2 theta + 2 sin theta)/(2 sin theta cos theta)`

      =`(2 sin theta (1+ sin theta))/(2 sin theta cos theta)`

      =`(1+sin theta)/cos theta`

      = RHS

 In the given question, we need to prove `((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2 = (1 - cos theta)/(1 + cos theta)`

Taking `sin theta` common from the numerator and the denominator of the L.H.S, we get

`((1 + sin theta - cos theta)/(1 + sin theta + cos theta))^2  = (((sin theta)(cosec theta + 1 -cot theta))/((sin theta)(cosec theta + 1 + cot theta)))^2`

`= ((1 + cosec theta - cot theta)/(1 + cosec theta + cot theta))^2`

Now, using the property  `1 + cot^2 theta = cosec^2 theta` we get

`((1 + cosec theta -  cot theta)/(1 + cosec theta + cot theta))^2 = (((cosec^2 theta - cot^2 theta) +cosec theta - cot theta)/(1 + cosec theta + cot theta))^2`

Using `a^2 - b^2 = (a + b)(a - b) we get

`(((cosec^2 theta - cot^2 theta)(cosec theta - cot theta))/(1 + cosec theta + cot theta))^2 = (((cosec theta - cot theta)(cosec theta + cot theta + 1))/(1 + cosec theta + cot theta))^2`

`= (cosec theta - cot theta)^2`

Using `cot theta = cos theta/sin theta` and `cosec = 1/sin theta` we get

`(cosec theta - cot theta)^2 = (1/sin theta - cos theta/sin theta)^2`

`= ((1 - cos theta)/sin theta)^2`

Now, using the property `sin^2 theta + cos^2 theta = 1` we get

`(1 - cos theta)^2/sin^2 theta = (1 - cos theta)/(1 - cos^2 theta)`

`= (1 - cos theta)^2/((1 + cos theta)(1 - cos theta))`

`= (1 - cos theta)/(1 + cos theta)`

Hence proved.

LHS = `sqrt((1/cos theta - 1)/(1/cos theta + 1)) + sqrt((1/cos theta +1)/(1/cos theta - 1))`

`= sqrt(((1 - cos theta)/cos theta)/((1+ cos theta)/cos theta)) + sqrt(((1 + cos theta)/cos theta)/((1 - cos theta)/cos theta)`

`= sqrt((1 - cos theta)/(1 + cos theta)) +sqrt((1 + cos theta)/(1 - cos theta))`

`= sqrt((1 - cos theta)/(1 + cos theta) xx (1 - cos theta)/(1 - cos theta)) + sqrt((1 + cos theta)/(1 - cos theta) xx (1 + cos theta)/(1 + cos theta))`

`= sqrt((1 - cos theta)^2/(1 - cos^2 theta)) + sqrt((1 + cos theta)^2/(1 - cos^2 theta))`

`=(1 - cos theta)/sin theta + (1 + cos theta)/sin theta`

`= (1 - cos theta + 1 + cos theta)/sin theta`

`= 2/sin theta`

= 2 cosec

LHS = `sqrt((1 - cos θ)/(1 + cos θ) xx (1 - cos θ)/(1 - cos θ))`

= `sqrt((1 - cos θ)^2/(1 - cos^2θ))`

= `(1 - cos θ)/(sqrt(1 - cos^2θ))` 

= `(1 - cos θ)/(sqrt(sin^2θ))`

= `(1 - cos θ)/(sin θ)`

= `(1)/(sin θ) - (cos θ)/(sin θ)`

= cosec θ − cot θ
= RHS
Hence proved.

We need to prove  `(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`

Here, we will first solve the LHS.

Now using `sec theta = 1/cos theta` and `tan theta = sin theta/cos theta`, we get

`(sec A - tan A)/(sec A + tan A) = (1/cos A - sin A/cos A)/(1/cos A + sin A/cos A)`

`= ((1 - sin A)/cos A)/((1 + sin A)/cos A)`

`= (1 - sin A)/(1 + sin A)`

Further, multiplying both numerator and denominator by 1 + sin A we get

`(1 - sin A)/(1 + sin A) = ((1 - sin A)/(1 + sin A))((1 + sin A)/(1 =  sin A))`

`= ((1 -sin A)(1 + sin A))/(1 + sin A)^2`

`= (1 - sin^2 A)/(1 + sin A)^2`

Now, using the property `cos^2 theta + sin^2 theta = 1`, we get

So,

`(1 - sin^2 A)/(1 + sin A)^2 = cos^2 A/(1 + sin A)^2`  = RHS.

Hence proved

Given that, tan A = n tan B and sin A = m sin B.

`=> n = tanA/tanB` and `m = sinA/sinB` 

∴ `(m^2 - 1)/(n^2 - 1) = ((sinA/sinB)^2 - 1)/((tanA/tanB)^2 - 1)`

= `(sin^2A/sin^2B - 1/1)/(tan^2A/(tan^2B) - 1)`

= `((sin^2A - sin^2B).tan^2B)/(sin^2B.(tan^2A - tan^2B))`

= `((sin^2A - sin^2B)/tan^2B)/((tan^2A - tan^2B)/sin^2B)`

= `((sin^2A - sin^2B)sin^2B)/((sin^2A/cos^2A-sin^2B/cos^2B)cos^2Bsin^2B)`

= `(sin^2A - sin^2B)/(((sin^2A.cos^2B - sin^2B.cos^2A)/(cos^2A.cos^2B)) cos^2B)`

= `((sin^2A - sin^2B)cos^2A)/(sin^2A.cos^2B - sin^2B.cos^2A)`

= `((sin^2A - sin^2B)cos^2A)/(sin^2A(1 - sin^2B) - sin^2B (1 - sin^2A))`

= `((sin^2A - sin^2B)cos^2A)/(sin^2A - sin^2A.sin^2B - sin^2B + sin^2B.sin^2A)`

= `((sin^2A - sin^2B)cos^2A)/(sin^2A -sin^2B)`

= cos² A

We have to prove `(1 - cos theta)/sin theta = sin theta/(1 + cos theta)`

We know that, `sin^2 theta + cos^2 theta = 1`

Multiplying both numerator and denominator by `(1 + cos theta)`, we have

`(1 - cos theta)/sin theta = ((1 - cos theta)(1 + cos theta))/(sin theta(1 + cos theta))`

`= (1 - cos^2 theta)/(sin theta(1 + cos theta))`

` = (sin^2 theta)/(sin theta(1 + cos theta))`

`= sin theta/(1 + cos theta)`

LHS = `(sec A)/(sec A + tan A - 1) + cos A/(cosec A + cot A - 1)`

= `(sin A)/(1/cos A + sin A/cos A - 1) + cos A/(1/sin A + cos A/sin A - 1)`

= `(sin A/(1 + sin A - cos A))/cos A + ((cos A)/(1 + cos A - sin A))/(sin A)`

= `(sin A.cos A)/(1 + sin A - cos A) + (sin A. cos A)/(1 + cos A - sin A)`

= `(sin A. cos A( 1 + cos A - sin A + 1 + sin A - cos A))/([ 1 + (sin A - cos A)][1 - (sin A - cos A)])`

= `(2sin A. cos A)/((1)^2 - (sin A - cos A)^2)`

= `(2sin A. cos A)/(1 - (sin^2 A + cos^2 A - 2 sin A.cos A))`

= `(2 sin A. cos A)/(1 - 1 + 2 sin A. cos A)`

= `2/2 = 1`

= RHS

Hence proved.

We have to prove `(1 + sin θ)/cos θ + cos θ/(1 + sin θ) = 2 sec θ`

We know that, `sin^2 θ + cos^2 θ = 1`

Multiplying the denominator and numerator of the second term by (1 − sin θ), we have

= `(1 + sin θ)/cos θ + cos θ/(1 + sin θ)`

`(1 + sin θ)/cos θ =  (cos θ(1 - sin θ))/((1 + sin θ)(1 - sin θ))`

`(1 + sin θ)/cos θ =  (cos θ (1 - sin θ))/(1-sin θ)`

= `(1 + sin θ)/cos θ + (cos θ(1 - sin θ))/cos^2 θ`

= `(1 + sin θ)/cos θ + (1 - sin θ)/cos θ`

= `(1 + sin θ +  1 - sin θ)/cos θ`

`= 2/cos θ`

= 2 sec θ

LHS = `(1 + sin θ)/cos θ + cos θ/(1 + sin θ)`

= `(( 1 + sin θ)^2 + cos^2 θ)/(cos θ(1 + sin θ))`

= `(1 + sin^2 θ + 2 sin θ + cos^2 θ)/(cos θ(1 + sin θ ))`

= `(1 + (sin^2θ + cos^2 θ) + 2 sin θ)/(cos θ(1 + sin θ))`

= `(1 + 1 + 2sin θ)/(cos θ(1 + sin θ))`

= `(2(1 + sin θ))/(cos θ(1 + sin θ))`

= 2 sec θ

Hence proved.

We know that `sec^2 theta - tan^2 theta = 1`

So,

`tan theta + 1/tan theta = (tan^2 theta + 1)/tan theta`

`= sec^2 theta/tan theta`

`= sec theta sec theta/tan theta`

`= sec theta = (1/cos theta)/(sin theta/cos theta)`

`= sec theta cosec theta`

L.H.S. = `(cos(90^circ - A))/(sin A)`

= `(sin A)/(sin A)`

= 1

R.H.S. = `(sin(90^circ - A))/(cos A)`

= `(cos A)/(cos A)`

= 1

∴ L.H.S. = R.H.S.

LHS = `(sec θ - tan θ)/(sec θ + tan θ )`

= `(sec θ - tan θ)/(sec θ + tan θ ) xx (sec θ - tan θ)/(sec θ - tan θ )`

= `(sec θ - tan θ)^2/(sec^2θ - tan^2θ )`

= `(sec^2θ + tan^2θ - 2sec θ.tan θ )/1`

= 1 + 2 tan²θ − 2 sec θ. tan θ

= R.H.S.
Hence proved.

In the given question, we need to prove `(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`

Here, we will first solve the LHS.

Now using `tan theta = sin theta/cos theta` and `cot theta = cos theta/sin theta` we get

`tan^2 A/(1 + tan^2 A) + cot^2 A/(1 + cot^2 A) = ((sin^2 A/cos^2 A))/((1 + sin^2 A/cos^2 A)) + ((cos^2 A/sin^2 A))/((1 + cos^2 A/sin^2 A))`

`= ((sin^2 A/cos^2 A))/(((cos^2 + sin^2 A)/cos^2 A)) + ((cos^2 A/sin^2 A))/(((sin^2 A + cos^2 A)/sin^2 A))`

`= ((sin^2 A/cos^2 A))/((1/cos^2 A)) + ((cos^2 A/sin^2 A))/((1/(sin^2  A)))`    (using `sin^2 theta + cos^2 theta = 1`)

On further solving by taking the reciprocal of the denominator, we get,

`(sin^2 A/cos^2 A)/(1/cos^2 A) + (cos^2 A/sin^2 A)/(1/sin^2 A) = ((sin^2 A)/(cos^2 A)) (cos^2 A/1) + (cos^2 A/sin^2 A)(sin^2 A/1)`

`= sin^2 A + cos^2 A`        (Using `sin^2 theta + cos^2 theta = 1`)

= 1

Hence proved.

LHS = mn

`= (cosec theta + cot theta) (cosec theta - cot theta)`

`= cosec^2 theta - cot^2 theta`

= 1    [∵ `(a + b)(a - b) = a^2 - b^2 cosec^2 theta - cot^2 theta = 1`]

= RHS Proved

In the given question, we need to prove `1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`

Here, we will first solve the L.H.S.

Now using `sec theta = 1/cos theta` and `tan theta = sin theta/cos theta`, we get

`1/(sec A +  tan A) - 1/cos A  = 1/(1/cos A + sin A/cos A) - (1/cos A)`

`= 1/(((1 + sin A)/cos A)) - (1/cos A)`

`= (cos A/(1 + sin A)) - (1/cos A)`

`= (cos^2 A - (1 + sin A))/((1 + sin A)(cos A))`

On further solving, we get

`(cos^2 A -(1 + sin A))/((1 + sin A)(cos A)) = (cos^2 A - 1 - sin A)/((1 +  sin A)(cos A))`

`= (-sin^2 A - sin A)/((1 + sin A)(cos A))`    (Using `sin^2 theta = 1 - cos^2 theta)`

`= (-sin A(sin A + 1))/((1 + sin A)(cos A))`

`= (-sin A)/cos A`

= − tan A

Similarly, we solve the R.H.S.

`((1 - sin A) - cos^2 A)/((cos A)(1 - sin^2 A)) = (1 - sin A - cos^2 A)/((cos A)(1 - sin A))`

`= (sin^2 A - sin A)/((cos A)(1 - sin A))`   (Using `sin^2 theta = 1- cos^2 theta`) 

`= (-sin A(1 - sin A))/((cos A)(1 - sin A))`

`= (-sin A)/cos A`

= − tan A

So, L.H.S = R.H.S

Hence proved.

We have to prove `sin theta/(1 - cos theta) = cosec theta + cot theta`

We know that `sin^2 theta + cos^2 theta = 1`

`sin theta/(1 - cos theta) = (sin theta (1 + cos theta))/(1 - cos^2 theta)`

`= (sin theta (1 +  cos theta))/(1 - cos^2 theta)`

`= (sin theta (1 + cos theta))/(sin^2 theta)`

`= (1 + cos theta)/sin theta`

`= 1/sin theta + cos theta/sin theta`

`= cosec theta  + cot theta`

We know that

sec² θ − tan² θ = 1

cosec² θ − cot² θ = 1

So,

(sec² θ − 1)(cosec² θ − 1) = tan² θ × cot² θ

= (tan θ × cot θ)

= `(tan θ xx 1/tan θ)^2`

= (1)²

= 1

We need to prove `sec^6 theta = tan^6 theta + 3 tan^2 theta sec^2 theta + 1`

Solving the L.H.S, we get

`sec^6 theta = (sec^2 theta)^3`

`= (1 + tan^2 theta)^3`

Further using the identity `(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2`, we get

`(1 + tan^2 theta)^3 = 1 + tan^6 theta + 3(1)^2 (tan^2 theta) + 3(1)(tan^2 theta)^2`

`= 1 + tan^6 theta + 3 tan^2 theta + 3 tan^4 theta`

`= 1 + tan^6 theta + 3 tan^2 theta + 3 tan^4 theta`

`= 1 + tan^6 theta + 3 tan^2 theta (1 + tan^2 theta)`

`= 1 + tan^6 theta + 3 tan^2 theta sec^2 theta`   (using `1 + tan^2 theta = sec^2 theta`)

Hence proved.

We need to prove `tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`

Now using cot θ = `1/tan θ` in the LHS, we get

`tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = tan θ/(1 - 1/tan θ) + (1/tan θ)/(1 - tan θ)`

`= tan θ/(((tan θ - 1)/tan θ)) + 1/(tan θ(1 - tan θ))`

`= (tan θ)/(tan θ  - 1)(tan θ) + 1/(tan θ(1 - tan θ)`

`= tan^2 θ/(tan θ - 1) - 1/(tan θ(tan θ - 1))`

`= (tan^3 θ - 1)/(tan θ(tan θ - 1))`

Further using the identity `a^3 - b^3 = (a - b)(a^2 + ab + b^2)`, we get

`(tan^3 θ - 1)/(tan(tan θ - 1)) = ((tan θ - 1)(tan^2 θ + tan θ + 1))/(tan θ (tan θ - 1))`

`= (tan^2 θ + tan θ + 1)/(tan θ)`

`= tan^2 θ/tan θ+ tan θ/tan θ + 1/tan θ`

= tan θ + 1 + cot θ

Hence `tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`

We have to prove  `(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`

We know that, sin² θ + cos² θ = 1

Multiplying both numerator and denominator by  (1 − sin θ), we have

`(1 - sin θ)/(1 + sin θ) = ((1 - sin θ)(1 -  sin θ))/((1 + sin θ)(1 - sin θ))`

`= (1 - sin θ)^2/(1 - sin^2 θ)`

`= ((1 - sin θ)/cos θ)^2`

`= (1/cos θ - sin θ/cos θ)^2`

`= (sec θ - tan θ)^2`

We have to prove `(1 + tan^2 theta)(1 - sin theta)(1 + sin theta) = 1`

We know that

`sin^2 theta + cos^2 theta = 1`

`sec^2 theta - tan^2 theta = 1`

So

`(1 + tan^2 theta)(1 - sin theta) = (1 + tan^2 theta){(1 - sin theta)(1 + sin theta)}`

` = (1 + tan^2 theta)(1 - sin^2 theta)`

`= sec^2 theta cos^2 theta`

` = 1/cos^2 theta cos^2 theta`

= 1

`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`

Consider the LHS = `(1 + cos θ + sin θ)/(1 + cos θ - sin θ)`

`= ((1 + cos θ + sin θ)/(1 + cos θ - sin θ))((1 + cos θ + sin θ)/(1 + cos θ + sin θ))`

`= (1 + cos θ + sin θ)^2/((1 + cos θ)^2 sin^2 θ)`

`= (2 + 2(cos θ + sin θ + sin θ cos θ))/(2 cos^2 θ+ 2 cos θ)`

`= (2(1 + cos θ)(1 + sin θ))/(2 cos θ (1 + cos θ))`

`= (1 + sin θ)/cos θ`

= RHS

Hence proved

LHS = `(sin θ/cos θ + sin θ)/(sin θ/cos θ - sin θ)`

= `(sin θ (1/cos θ + 1))/(sin θ (1/cos θ - 1))`

= `(sec θ + 1)/(sec θ - 1)`

= RHS

Hence proved.

LHS = `sqrt((1 + sin A)/(1 - sin A))`

= `sqrt((1 + sin A)/(1 - sin A) xx (1 + sin A)/(1 + sin A)`

= `sqrt((1 + sin A)^2/(1 - sin^2 A))`

= `sqrt((1 + sin A)^2/cos^2 A)`

= `(1 + sin A)/cos A`

= sec A + tan A = RHS

L.H.S. = `"cosec"  θ xx sqrt(1 - cos^2θ)`

= `"cosec"  θ xx sqrt(sin^2θ)`   ...`[(∵ sin^2θ + cos^2θ = 1),(therefore 1 - cos^2θ = sin^2θ)]`

= cosec θ × sin θ

= 1   ...[∵ sin θ × cosec θ = 1]

= R.H.S.

LHS = `sqrt(( secθ - 1)/(secθ + 1)) + sqrt((secθ + 1)/(secθ - 1))` 

= `(sqrt( secθ - 1) sqrt( secθ - 1) + sqrt( secθ + 1)sqrt( secθ + 1))/(sqrt(secθ - 1)sqrt(secθ + 1))`

= `((sqrt( secθ - 1))^2 + (sqrt( secθ + 1))^2)/(sqrt(secθ - 1)sqrt(secθ + 1))`

= `(secθ - 1 + secθ + 1)/(sqrt(sec^2 - 1))`

= `(2secθ)/sqrt(tan^2θ)`

= `(2secθ)/(tanθ)`

= `(2 1/cosθ)/(sinθ/cosθ)`

= `(2 1/sinθ)`

= 2 cosecθ.

Given: x = h + a cos θ

x − h = a cos θ   ...(i)

y = k + b sin θ

y − k = b sin θ    ...(ii)

The given equation is

`((x - h)/a)^2 + ((y - k)/(b))^2 = 1`

LHS = `((a cos θ)/a)^2 + ((b sin θ)/b)^2 `     ...[Putting the values of (i) and (ii)]

= cos²θ + sin²θ

= 1

= RHS

Hence proved.

We have,

sinθ + sin² θ = 1

⇒ sinθ = 1 – sin² θ

⇒ sin θ = cos² θ    ......[∵ sin² θ +  cos² θ = 1]

(sinθ)² = (cos² θ)²

sin² θ = cos⁴ θ

= cos² θ + cos⁴ θ

= sin θ + sin² θ

cos² θ + cos⁴ θ = 1

L.H.S. = sin⁶A + cos⁶A

= (sin²A)³ + (cos²A)³   

= (1 – cos²A)³ + (cos²A)³    ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`

= 1 – 3cos²A + 3(cos²A)² – (cos²A)³ + cos⁶A   ...[∵ (a – b)³ = a³ – 3a²b + 3ab² – b³]

= 1 – 3 cos²A (1 – cos²A) – cos⁶A + cos⁶A

= 1 – 3 cos²A sin²A

= R.H.S.

∴ sin⁶A + cos⁶A = 1 – 3sin²A . cos²A

L.H.S. = (1 – cos²A) . sec²B + tan²B (1 – sin²A)

= `sin^2A * 1/(cos^2B) + (sin^2B)/(cos^2B) (1 - sin^2A)`   ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`

= `(sin^2A)/(cos^2B) + (sin^2B)/(cos^2B) - (sin^2A sin^2B)/(cos^2B)`

= `(sin^2A)/(cos^2B) - (sin^2A sin^2B)/(cos^2B) + (sin^2B)/(cos^2B)`

= `(sin^2A)/(cos^2B) (1 - sin^2B) + tan^2B`

= `(sin^2A)/(cos^2B) (cos^2B) + tan^2B`

= sin²A + tan²B

= R.H.S.

∴ (1 – cos²A) . sec²B + tan²B (1 – sin²A) = sin²A + tan²B

L.H.S. = \[\boxed{\text{tan}^2θ - \text{sin}^2θ}\] 

= \[\boxed{\text{tan}^2θ} \left(1 - \frac{\text{sin}^2θ}{\text{tan}^2θ}\right)\]

= \[\tan^2\theta\left(1-\frac{\boxed{\sin^2\theta}}{\frac{\sin^2\theta}{\cos^2\theta}}\right)\]

= \[\tan^{2}\theta\left(1-\frac{\sin^{2}\theta}{1}\times\frac{\cos^{2}\theta}{\boxed{\sin^{2}\theta}}\right)\]

= \[\text{tan}^2θ \left(1 - \boxed{\text{cos}^2θ}\right)\]

= \[\text{tan}^2θ × \boxed{\text{sin}^2θ}\]   ...[1 – cos²θ = sin²θ]

= R.H.S.

L.H.S. = `(sin θ + "cosec"  θ)/(sin θ)`

= `(sin θ)/(sin θ) + ("cosec"  θ)/(sin θ)`

= 1 + cosec θ × cosec θ   ...`[∵ "cosec"  θ = 1/(sin θ)]`

= 1 + cosec²θ

= 1 + 1 + cot²θ   ...[∵ 1 + cot²θ = cosec²θ]

= 2 + cot²θ

= R.H.S.

∴ `(sin θ + "cosec"  θ)/(sin θ) = 2 + cot^2θ`

L.H.S. = cot²θ × sec²θ

= `(cos^2θ)/(sin^2θ) xx 1/(cos^2θ)`

= `1/(sin^2θ)`

= cosec²θ

= 1 + cot²θ   ...[∵ 1 + cot²θ = cosec²θ]

= R.H.S.

∴ cot²θ × sec²θ = cot²θ + 1

L.H.S. = `(1 + sin θ)/(1 - sin θ)`

= `((1 + sinθ)/(cosθ))/((1 - sinθ)/(cosθ))`   ...[Dividing numerator and denominator by cos θ]

= `(1/cosθ + (sinθ)/(cosθ))/(1/cosθ - (sinθ)/(cosθ)`

= `(secθ + tanθ)/(secθ - tanθ)`

= `(secθ + tanθ)/(secθ - tanθ) xx (secθ + tanθ)/(secθ + tanθ)`   ...[On rationalising the denominator]

= `(secθ + tanθ)^2/(sec^2θ - tan^2θ)`

= `(secθ + tanθ)^2/1`   ...`[(∵ 1 + tan^2θ = sec^2θ),(∴ sec^2θ - tan^2θ = 1)]`

= (sec θ + tan θ)²

= R.H.S.

∴ `(1 + sinθ)/(1 - sinθ) = (sec θ + tan θ)^2` 

L.H.S. = `(cot A)/(1 - cot A) + (tan A)/(1 - tan A)`

= `(cot A)/(1 - 1/(tan A)) + (tan A)/(1 - tan A)`

= `(cot A)/((tan A - 1)/(tan A)) + (tan A)/(1 - tan A)`

= `(cot A tan A)/(tan A - 1) + (tan A)/(1 - tan A)`

= `1/(tan A - 1) + (tan A)/(1 - tan A)`   ...[∵ cot A tan A = 1]

= `- 1/(1 - tan A) + (tan A)/(1 - tan A)`

= `- (1/(1 - tan A) - (tan A)/(1 - tan A))`

= `-((1 - tan A)/(1 - tan A))`

= –1

= R.H.S.

∴ `(cot A)/(1 - cot A) + (tan A)/(1 - tan A) = -1`

sin⁶A + cos⁶A = (sin²A)³ + (cos²A)³

 = (1 – cos²A)³ + (cos²A)³    ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`

= 1 – 3 cos²A + 3(cos²A)² – (cos²A)³ + cos⁶A   ...[∵ (a – b)³ = a³ – 3a²b + 3ab² – b³]

= 1 – 3 cos²A(1 – cos²A) – cos⁶A + cos⁶A

= 1 – 3 cos²A sin²A

sin⁴A + cos⁴A = (sin²A)² + (cos²A)²

= (1 – cos²A)² + (cos²A)²

= 1 – 2 cos²A + (cos²A)² + (cos²A)²    ...[∵ (a – b)² = a² – 2ab + b²]

= 1 – 2 cos²A + 2 cos⁴A

= 1 – 2 cos²A(1 – cos²A)

= 1 – 2 cos²A sin²A

L.H.S. = 2(sin⁶A + cos⁶A) – 3(sin⁴A + cos⁴A) + 1

= 2(1 – 3 cos²A sin²A) – 3(1 – 2 cos²A sin²A) + 1

= 2 – 6 cos²A sin²A – 3 + 6 cos²A sin²A + 1

= 0

= R.H.S.

∴ 2(sin⁶A + cos⁶A) – 3(sin⁴A + cos⁴A) + 1 = 0

L.H.S = `(1 + sin B)/(cos B) + (cos B)/(1 + sin B)`

= `((1 + sin B)^2 + cos^2B)/(cos B(1 + sin B))`

= `(1 + 2 sin B + sin^2B + cos^2B)/(cos B(1 + sin B))`   ...[∵ (a + b)² = a² + 2ab + b²]

= `(1 + 2 sin B + 1)/(cos B(1 + sin B))`   ...[∵ sin²B + cos²B = 1]

= `(2 + 2 sin B)/(cos B(1 + sin B))`

= `(2(1 + sin B))/(cos B(1 + sin B))`

= `2/(cos B)`

= 2 sec B

= R.H.S.

∴ `(1 + sin B)/(cos B) + (cos B)/(1 + sin B) = 2 sec B`

cosec A – sin A = p   ...[Given]

∴ `1/(sin A) - sin A = p`

∴ `(1 - sin^2A)/(sin A) = p`

∴ `(cos^2A)/(sin A) = p`   ...(i) `[(∵ sin^2A + cos^2A = 1),(∴ 1 - sin^2A = cos^2A)]`

sec A – cos A = q   ...[Given]

∴ `1/(cos A) - cos A = q`

∴ `(1 - cos^2A)/(cos A) = q`

∴ `(sin^2A)/(cos A) = q`   ...(ii) `[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`

L.H.S. = `(p^2q)^(2/3) + (pq^2)^(2/3)`

= `[((cos^2A)/(sin A))^2 ((sin^2A)/(cos A))]^(2/3) + [((cos^2A)/(sin A))((sin^2A)/(cos A))^2]^(2/3)`   ...[From (i) and (ii)]

= `((cos^4A)/(sin^2A) xx (sin^2A)/(cos A))^(2/3) + ((cos^2A)/(sin A) xx (sin^4A)/(cos^2A))^(2/3)`

= `(cos^3A)^(2/3) + (sin^3A)^(2/3)`

= cos²A + sin²A

= 1

= R.H.S.

∴ `(p^2q)^(2/3) + (pq^2)^(2/3) = 1`

L.H.S. = `sqrt((1 + cos A)/(1 - cos A))`

= `sqrt((1 + cos A)/(1 - cos A) xx (1 + cos A)/(1 + cos A))`   ...[On rationalising the denominator]

= `sqrt((1 + cos A)^2/(1 - cos^2 A))`

= `sqrt((1 + cos A)^2/(sin^2 A)`   ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`

= `(1 + cos A)/(sin A)`

= `1/(sin A) + (cos A)/(sin A)`

= cosec A + cot A

= R.H.S.

∴ `sqrt((1 + cos A)/(1 - cos A)) = "cosec"  A + cot A`

L.H.S. = sec²θ – cos²θ

= sec²θ – (1 – sin²θ)   ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 - sin^2θ = cos^2θ)]`

= sec²θ – 1 + sin²θ

= tan²θ + sin²θ   ...`[(∵ 1 + tan^2θ = sec^2θ),(∴ tan^2θ = sec^2θ - 1)]`

= R.H.S.

∴ sec²θ – cos²θ = tan²θ + sin²θ

L.H.S. = cosec θ – cot θ

= `1/(sin θ) - (cos θ)/(sin θ)`

= `(1 - cos θ)/(sin θ)`

= `(1 - cos θ)/(sin θ) xx (1 + cos θ)/(1 + cos θ)`   ...[On rationalising the numerator]

= `(1 - cos^2θ)/(sinθ(1 + cosθ))`

= `(sin^2θ)/(sinθ(1 + cosθ))`   ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 - cos^2θ = sin^2θ)]`

= `(sin θ)/(1 + cos θ)`

= R.H.S.

∴ cosec θ – cot θ = `(sin θ)/(1 + cos θ)`

L.H.S. = `(cot A + "cosec" A - 1)/(cot A - "cosec" A + 1)`

= `(cot A + "cosec" A - ("cosec"^2A - cot^2A))/(cot A - "cosec" A + 1)`   ...`[(∵ 1 + cot^2A = "cosec"^2A),(∴ "cosec"^2A - cot^2A = 1)]`

= `(cot A + "cosec" A - ("cosec" A + cot A)("cosec" A - cot A))/(cot A - "cosec" A + 1)`   ...[∵ a² – b² = (a + b) (a – b)]

= `((cot A + "cosec" A)(1 - "cosec" A + cot A))/(cot A - "cosec" A + 1)`

= cot A + cosec A

= `(cos A)/(sin A) + 1/(sin A)`

= `(cos A + 1)/(sin A)`

= R.H.S.

∴ `(cot A + "cosec" A - 1)/(cot A - "cosec" A + 1) = (1 + cos A)/(sin A)`

L.H.S. = sec²A – cosec²A

= `1/(cos^2A) - 1/(sin^2A)`

= `(sin^2A - cos^2A)/(cos^2A*sin^2A)`

= `(sin^2A - (1 - sin^2A))/(sin^2A*cos^2A)`   ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - sin^2A = cos^2A)]`

= `(sin^2A - 1 + sin^2A)/(sin^2A*cos^2A)`

= `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`

= R.H.S.

∴ `sec^2A - "cosec"^2A = (2sin^2A - 1)/(sin^2A*cos^2A)`

`cos A = (2sqrt(m))/(m + 1)`   ...[Given]

We know that,

sin²A + cos²A = 1

∴ `sin^2A + ((2sqrt(m))/(m + 1))^2 = 1`

∴ `sin^2A + (4m)/(m + 1)^2 = 1`

∴ `sin^2A = 1 - (4m)/(m + 1)^2`

= `((m + 1)^2 - 4m)/(m + 1)^2`

= `(m^2 + 2m + 1 - 4m)/(m + 1)^2`   ...[∵ (a + b)² = a² + 2ab + b²]

= `(m^2 - 2m + 1)/(m + 1)^2`

∴ `sin^2A = (m - 1)^2/(m + 1)^2`   ...[∵ a² – 2ab + b² = (a – b)²]

∴ `sin A = (m - 1)/(m + 1)`   ...[Taking square root of both sides]

Now, `"cosec"  A = 1/(sin A)`

= `1/((m - 1)/(m + 1))`

∴ `"cosec"  A = (m + 1)/(m - 1)`

L.H.S. = `(sec A)/(tan A + cot A)`

= `(sec A)/((sin A)/(cos A) + (cos A)/(sin A))`

= `(sec A)/((sin^2A + cos^2A)/(cosA sinA))`

= `(sec A)/(1/(cosA sinA))`   ...[∵ sin²A + cos²A = 1]

= sec A cos A sin A

= `1/(cos A) xx cos A sin A`

= sin A

= R.H.S.

∴ `(sec A)/(tan A + cot A) = sin A`

Given a cos θ – b sin θ = c

Squaring on both sides

(a cos θ – b sin θ)² = c²

a² cos² θ + b² sin² θ – 2 ab cos θ sin θ = c²

a² (1 – sin² θ) + b² (1 – cos² θ) – 2 ab cos θ sin θ = c²

a² – a² sin² θ + b² – b² cos² θ – 2 ab cos θ sin θ = c² 

– a² sin² θ – b²cos² θ – 2 ab cos θ sin θ  = – a² – b² + c²

a² sin² θ + b² cos² θ + 2 ab cos θ sin θ = a² + b² – c²

(a sin θ + b cos θ)² – a² + b² – c² 

a sin θ + b cos θ = `±  sqrt(a^2 + b^2 - c^2)`

Hence, it is proved.

In the given question, we need to prove `1 + cot^2 theta/(1 + cosec theta) = cosec theta`

Using `cot theta = cos theta/sin theta` and `cosec theta = 1/sin theta` We get

`1 + cot^2 theta/(1 +  cosec theta) = (1 = cosec theta +  cot^2 theta)/(1 + cosec theta)`

`= ((1 + 1/sin theta + cos^2 theta/sin^2 theta))/((1 + 1/sin theta))`

` = (((sin^2 theta + sin theta + cos^2 theta)/sin^2 theta))/(((sin theta + 1)/sin theta))`

Further, using the property `sin^2 theta + cos^2 theta = 1`

We get

`((sin^2 theta + sin theta + cos^2 theta)/sin^2 theta)/((sin theta + 1)/sin theta) = ((1 + sin theta)/sin^2 theta)/((sin theta + 1)/sin theta)`

`= (1 + sin theta/sin^2 theta)((sin theta)/(1 + sin theta))`

`= 1/sin theta`

`= cosec theta`

Hence proved.

We have to prove the following identity

`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta  + cot theta`

Consider the LHS = `(cos theta - sin theta + 1)/(cos theta + sin theta - 1)`

`= (cos theta - sin theta  +  1)/(cos theta +  sin theta  - 1) xx (cos theta +  sin theta +  1)/(cos theta +  sin theta + 1)`

`= ((cos theta + 1)^2 - (sin theta)^2)/((cos theta +  sin theta)^2 - (1)^2)`

`= (cos^2 theta +  1 +  2 cos theta - sin^2 theta)/(cos^2 theta +  sin^2 theta + 2 cos theta sin theta - 1)`

`= (cos^2 theta + 1 +  2 cos theta - (1 -  cos^2 theta))/(1 +  2 cos theta sin theta  - 1)`

`= (2 cos^2 theta + 2 cos theta)/(2 cos theta sin theta)`

`= (2 cos^2 theta +  2 cos theta)/(2 cos theta sin theta)`

`= (2 cos theta(cos theta + 1))/(2 cos theta sin theta)`

`= (cos theta + 1)/sin theta`

`= cos theta/sin theta + 1/sin theta`

`= cot theta + cosec theta`

= RHS

We have to prove `(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`

Consider the expression

L.H.S

`(1 + tan^2 theta)/(1 + cot^2 theta) = (1 + tan^2 theta)/(1 + 1/(tan^2 theta))`

= `(1 +tan^2 theta)/((tan^2 theta + 1)/tan^2 theta)`

`= tan^2 theta (1 + tan^2 theta)/(1 + tan^2 theta)`

`= tan^2 theta` 

= R.H.S

Again, we have 

L.H.S

`((1 - tan theta)/(1 - cot theta))^2 = ((1 - tan theta)/(1 - 1/(tan theta)))^2`

`= ((1 - tan theta)/((tan theta - 1)/tan theta))^2`

`=[(tantheta(1-tantheta))/-(1-tantheta)]^2`

`=(-tantheta)^2=tan^2theta`

= R.H.S

Given: 1 + sin² θ = 3 sin θ cos θ

Dividing L.H.S and R.H.S equations with sin²θ,

We get, 

`(1 + sin^2 theta)/(sin^2 theta) = (3 sin theta cos theta)/(sin^2 theta)`

`\implies 1/(sin^2 theta) + 1 = (3 cos theta)/sintheta`

cosec² θ + 1 = 3 cot θ

Since, cosec² θ – cot² θ = 1 

`\implies` cosec² θ = cot² θ + 1

`\implies` cot² θ + 1 + 1 = 3 cot θ

`\implies` cot² θ + 2 = 3 cot θ

`\implies` cot² θ – 3 cot θ + 2 = 0

Splitting the middle term and then solving the equation,

`\implies` cot² θ – cot θ – 2 cot θ + 2 = 0

`\implies` cot θ(cot θ – 1) – 2(cot θ + 1) = 0

`\implies` (cot θ – 1)(cot θ – 2) = 0

`\implies` cot θ = 1, 2

Since,

tan θ = `1/cot θ`

tan θ = `1, 1/2`

Hence proved.

Given, 1 + sin² θ = 3 sin θ cos θ

On dividing by sin² θ on both sides, we get

`1/(sin^2θ) + 1 = 3 cot θ`   ...`[∵ cot θ = cos θ/sin θ]`

⇒ cosec² θ + 1 = 3 cot θ

⇒ 1 + cot² θ + 1 = 3 cot θ

⇒ cot² θ – 3 cot θ + 2 = 0

⇒ cot² θ – 2 cot θ – cot θ + 2 = 0

⇒ cot θ (cot θ – 2) – 1(cot θ – 2) = 0

⇒ (cot θ – 2) (cot θ – 1) = 0

⇒ cot θ = 1 or 2

tan θ = 1 or `1/2`

Hence proved.

L.H.S. = (cosec A – sin A) (sec A – cos A) (tan A + cot A) 

= `(1/sinA - sinA)(1/cosA - cosA)(1/tanA + tanA)`

= `((1 - sin^2A)/sinA)((1 - cos^2A)/cosA)(sinA/cosA + cosA/sinA)`

= `(cos^2A/sinA)(sin^2A/cosA)((sin^2A + cos^2A)/(sinA.cosA))`

= `(cos^2A/sinA)(sin^2A/cosA)((1)/(sinA.cosA))`

= `(cos^2A sin^2A)/((sinA .cosA)(sinA.cosA ))`

= `(cos^2A sin^2A)/(sin^2A cos^2A)`

= 1

= R.H.S.

L.H.S. = (cos A + sin A)² + (cos A – sin A)²

= cos² A + sin² A + 2 cos A . sin A + cos² A + sin² A – 2 cos A . sin A

= 2 sin² A + 2 cos² A

= 2(sin² A + cos² A)   ...(∵ sin² A + cos² A = 1)

= 2 × 1    

= 2

= R.H.S.

L.H.S. = `(cot A - 1)/(2 - sec^2A)`

= `(((1 - tan A))/tan A)/(1 + 1 - sec^2A)` 

= `(((1 - tan A))/tan A)/(1 - tan^2A)` 

=  `(((1 - tan A))/(tan A))/((1 + tan A)(1 - tan A))`

= `(1/tan A)/(1 + tan A)`

= `cot A/(1 + tan A)`

= R.H.S.

Hence proved.

LHS = (sin θ + cos θ)(cosec θ – sec θ)

= `(sin θ + cos θ)(1/sin θ - 1/cos θ)`

= `(sin θ + cos θ)((cos θ - sin θ)/(sin θ * cos θ))`

= `(cos^2θ - sin^2θ)/(sinθ * cosθ)`

= `(1 - 2sin^2θ)/(sinθ*cosθ)`

= `1/(sinθ * cosθ) - (2 sin^2θ)/(sinθ * cosθ)`

= `cosec θ · sec θ - (2 sin^2 θ)/(sin θ * cos θ)`

= cosec θ · sec θ – 2 tan θ

= RHS

Hence proved.

L.H.S. = `1/("cosec"  θ - cot θ)`

= `1/("cosec"  θ - cot θ) xx ("cosec"  θ + cot θ)/("cosec"  θ + cot θ)`   ...[On rationalising the denominator]

= `("cosec"  θ + cot θ)/("cosec"^2θ - cot^2θ)`   ...[∵ (a – b)(a + b) = a² – b²]

= `("cosec"  θ + cot θ)/1`   ...`[(∵ 1 + cot^2θ = "cosec"^2θ),(∴ "cosec"^2θ - cot^2θ = 1)]`

= cosec θ + cot θ = R.H.S.

∴ `1/("cosec"  θ - cot θ) = "cosec"  θ + cot θ`

L.H.S. = sin⁴A – cos⁴A 

= (sin²A)² – (cos²A)²

= (sin²A + cos²A)(sin²A – cos²A)   ...[∵ a² – b² = (a + b)(a – b)]

= (1)(sin²A – cos²A)   ...[∵ sin²A + cos²A = 1]

= sin²A – cos²A

= (1 – cos²A) – cos²A   ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`

= 1 – 2 cos²A

= R.H.S.

∴ sin⁴A – cos⁴A = 1 – 2 cos²A

L.H.S. = `(1 + sec A)/(sec A)`

= `1/(sec A) + (sec A)/(sec A)`

= cos A + 1

= `(1 + cos A) xx (1 - cos A)/(1 - cos A)`

= `(1 - cos^2A)/(1 - cosA)`

= `(sin^2A)/(1 - cosA)`   ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`

= R.H.S.

∴ `(1 + sec A)/(sec A) = (sin^2A)/(1 - cosA)`

L.H.S. = sin²A . tan A + cos²A . cot A + 2 sin A . cos A

= `sin^2A * (sin A)/(cos A) + cos^2A * (cos A)/(sin A) + 2 sin A * cos A`

= `(sin^3A)/(cos A) + (cos^3A)/(sin A) + 2 sin A * cos A`

= `(sin^4A + cos^4A + 2 sin^2A cos^2A)/(sinA cosA)`

= `(sin^2A + cos^2A)^2/(sinA cosA)`   ...[∵ a² + b² + 2ab = (a + b)²]

= `1^2/(sinA cosA)`   ...[∵ sin²A + cos²A = 1]

=  `1/(sinA cosA)`  

= `(sin^2A + cos^2A)/(sinA cosA)`   ...[∵ 1 = sin²A + cos²A]

= `(sin^2A)/(sinA cosA) + (cos^2A)/(sinA cosA)`

= `(sin A)/(cos A) + (cos A)/(sin A)`

= tan A + cot A

= R.H.S.

∴ sin²A . tan A + cos²A . cot A + 2 sin A . cos A = tan A + cot A

L.H.S. = sec²θ – cos²θ

= 1 + tan²θ – cos²θ   ...[∵ 1 + tan²θ = sec²θ]

= tan²θ + (1 – cos²θ)

= tan²θ + sin²θ   ...`[(∵ sin^2θ +cos^2θ = 1),(∴ 1 - cos^2θ = sin^2θ)]`

= R.H.S.

∴ sec²θ – cos²θ = tan²θ + sin²θ

L.H.S. = `(tan(90 - θ) + cot(90 - θ))/("cosec"  θ)`

= `1/("cosec"  θ)(cot θ + tan θ)`   ...`[(∵ tan(90 - θ) = cot θ),(cot(90 - θ) = tan θ)]`

= sin θ (cot θ + tan θ)

= `sin θ ((cos θ)/(sin θ) + (sin θ)/(cos θ))`

= `sin θ ((cos^2θ + sin^2θ)/(sinθ cosθ))`

= `sin θ (1/(sin θ cos θ))`   ...[∵ sin²θ + cos²θ = 1]

= `1/(cos θ)`

= sec θ

= R.H.S.

∴ `(tan(90 - θ) + cot(90 - θ))/("cosec"  θ) = sec θ`

L.H.S. = `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1)` 

= `(sin θ)/(1/cos θ + 1) + (sin θ)/(1/(cos θ) - 1`

= `(sin θ)/((1 + cos θ)/(cos θ)) + (sin θ)/((1 - cos θ)/(cos θ))`

= `(sin θ cos θ)/(1 + cos θ) + (sin θ cos θ)/(1 - cos θ)`

= `sin θ cos θ (1 /(1 + cos θ) + 1/(1 - cos θ))`

= `sin θ cos θ [(1 - cos θ + 1 + cos θ)/((1 + cos θ)(1 - cos θ))]`

= `sin θ cos θ (2/(1 - cos^2θ))`   ...[∵ (a + b)(a – b) = a² – b²]

= `sin θ cos θ xx 2/(sin^2θ)`   ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 - cos^2θ = sin^2θ)]`

= `2 xx (cos θ)/(sin θ)`

= 2 cot θ

= R.H.S.

∴ `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1) = 2 cot θ`

L.H.S. = `(sin^2θ)/(cos θ) + cos θ`

= `(sin^2θ + cos^2θ)/(cos θ)`

= `1/(cos θ)`   ...[∵ sin²θ + cos²θ = 1]

= sec θ

= R.H.S.

∴ `(sin^2θ)/(cos θ) + cos θ = sec θ`

L.H.S. = \[\boxed{\text{cot} \phantom{.} θ + \text{tan} \phantom{.}θ}\]

= \[\frac{\boxed{\text{cos}\phantom{.}θ}}{\text{sin}\phantom{.}θ} + \frac{\text{sin}\phantom{.}θ}{\text{cos}\phantom{.}θ}\]

= \[\frac{\text{cos}^2θ + \text{sin}^2θ}{\boxed{\text{sin}θ.\text{cos}θ}}\]

= `1/(sinθ.cosθ)`   ...[cos²θ + sin²θ = \[\boxed{1}\]]

= \[\frac{1}{\text{sin}θ} \times \frac{1}{\boxed{\text{cos}θ}}\]

= \[\boxed{\text{cosec} \phantom{.}θ \times \text{sec} \phantom{.}θ}\]

= R.H.S.

sin θ + cos θ = `sqrt(3)`

Squaring on both sides:

(sin θ + cos θ)² = `(sqrt(3))^2`

sin² θ + cos² θ + 2 sin θ cos θ = 3

1 + 2 sin θ cos θ = 3

2 sin θ cos θ = 3 – 1

2 sin θ cos θ = 2

∴ sin θ cos θ = 1

L.H.S = tan θ + cot θ

= `sin theta/cos theta + cos theta/sin theta`

= `(sin^2 theta + cos^2 theta)/(sin theta cos theta)`

= `1/(sin theta cos theta)`

= `1/1`   ...(sin θ cos θ = 1)

= 1 = R.H.S.

⇒ tan θ + cot θ = 1

L.H.S = R.H.S

L.H.S. = cot²θ – tan²θ

= (cosec²θ – 1) – (sec²θ – 1)   ...`[(∵ tan^2θ = sec^2θ - 1),(cot^2θ = "cosec"^2θ - 1)]`

= cosec²θ – 1 – sec²θ + 1

= cosec²θ – sec²θ

= R.H.S.

∴ cot²θ – tan²θ = cosec²θ – sec²θ 

`(cot A)/(1 - tan A) + (tan A)/(1 - cot A)`

= `((cos A)/(sin A))/(1 - (sin A)/(cos A)) + ((sin A)/(cos A))/(1 - (cos A)/(sin A))`

= `((cos A)/(sin A))/((cos A  -  sin A)/(cos A)) + ((sin A)/(cos A))/((sin A  -  cos A)/(sin A))`

= `(cos A)/(sin A) xx (cos A)/(cos A - sin A) + (sin A)/(cos A) xx (sin A)/(sin A - cos A)`

= `(cos^2A)/(sin A(cos A - sin A)) + (sin^2A)/(cos A(sin A - cos A))`

= `1/(sin A - cos A) ((-cos^3A + sin^3A)/(sin A cos A))`

= `1/(sin A - cos A)((sin^3A - cos^3A)/(sin A cos A))`

= `1/(sin A - cos A) xx ((sin A - cos A)(sin^2A + sin A cos A + cos^2A))/(sin A cos A)`   ...[∵ a³ – b³ = (a – b)(a² + ab + b²)]

= `(sin^2A + sin A cos A + cos^2A)/(sin A cos A)`   ...(i)

= `(1 + sin A cos A)/(sin A cos A)`   ...[∵ sin²A + cos²A = 1]

= `1/(sin A cos A) + (sin A cos A)/(sin A cos A)`

= cosec A sec A + 1   ...(ii)

`(cot A)/(1 - tan A) + (tan A)/(1 - cot A)`

= `(sin^2A + sin A cos A + cos^2A)/(sin A cos A)`   ...[From (i)]

= `(sin^2A)/(sin A cos A) + (sin A cos A)/(sin A cos A) + (cos^2A)/(sin A cos A)`

= `(sin A)/(cos A) + 1 + (cos A)/(sin A)`

= tan A + 1 + cot A   ...(iii)

From (ii) and (iii), we get

`(cot A)/(1 - tan A) + (tan A)/(1 - cot A) = 1 + tan A + cot A = sec A  .  "cosec"  A + 1`

L.H.S. = \[\boxed{\text{sin}^4A - \text{cos}^4A}\] 

= (sin²A)² – (cos²A)²

 = \[{(\text{sin}^2A + \text{cos}^2A) (\boxed{\text{sin}^2A - \text{cos}^2A})}\]   ...[∵ a² – b² = (a + b)(a – b)]

= \[1(\boxed{\text{sin}^2A - \text{cos}^2A})\]   ...[∵ sin²A + \[\boxed{\text{cos}^2\text{A}}\] = 1]

= sin²A – cos²A

= \[\boxed{1 - \text{cos}^2A} - \text{cos}^2A\]   ...[sin²A = 1 – cos²A]

= \[\boxed{1 - 2\text{cos}^2A}\]

= R.H.S.

L.H.S. = sin θ (1 – tan θ) – cos θ (1 – cot θ)

= `sin θ (1 - (sin θ)/(cos θ)) - cos θ (1 - (cos θ)/(sin θ))`

= `sin θ - (sin^2θ)/(cosθ) - cos θ + (cos^2θ)/(sinθ)`

= `sin θ + (cos^2θ)/(sinθ) - (sin^2θ)/(cosθ) - cos θ`

= `(sin^2θ + cos^2θ)/(sinθ) - ((sin^2θ + cos^2θ)/(cosθ))`

= `1/(sinθ) - 1/(cosθ)`   ...[∵ sin²θ + cos²θ = 1]

= cosec θ – sec θ

= R.H.S.

∴ sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ

LHS = `sin theta/((cot theta + cosec  theta))- sin theta/(( cot theta - cosec theta))`

= `sin theta { ((cot theta - cosec  theta )-( cot  theta + cosec  theta ))/(( cot theta + cosec  theta )  ( cot  theta - cosec  theta ))}`

= `sin theta { (-2 cosec  theta)/(-1)}    (∵ cosec^2  theta - cot^2  theta  =1)`

= `sin theta . 2 cosec  theta`

= `sin theta xx2xx1/ sin theta`

= 2

= RHS

LHS= `(1+ cos  theta - sin^2 theta )/(sin theta (1+ cos theta)`

     =` ((1+ cos theta )- (1-cos^2 theta))/(sin theta(1+ cos theta))`

     =`(cos theta + cos^2 theta)/( sin theta ( 1+ cos theta))`

     =`(cos theta ( 1+ cos theta ))/ ( sin theta ( 1+ cos theta))`

     =`cos theta/ sin theta`

     = cot 𝜃
     = RHS
Hence, L.H.S. = R.H.S. 

Here, `( cosec theta + cot theta )/( cosec theta - cot theta)`

        = `((cosec theta + cot theta) ( cosec theta + cot theta ))/(( cosec theta - cot theta ) ( cosec theta + cot theta))`

        =` ((cosec theta + cot theta)^2)/(( cosec ^2 theta - cot^2 theta))`

       =`((cosec theta + cot theta )^2) /1`

      =`(cosec theta + cot theta )^2`

 Again , `( cosec theta + cot theta )^2`

     = ` cosec^2 theta + cot^2 theta + 2 cosec theta  cot theta `

     =` 1+cot^2 theta + cot^2 theta + 2 cosec theta  cot theta (∵ cosec^2 theta - cot^2 theta =1)`

    =` 1+2 cot^2 theta + 2 cosec theta  cot theta `

We have `(x/a sin theta - y/a cos theta ) =1`

Squaring both side, we have:

`(x/a sin theta - y/b cos theta )^2 = (1)^2`

⇒ `(x^2/a^2 sin^2 theta + y^2/b^2 cos^2 theta - 2 x/a xx y/b sin theta cos theta ) = 1    .....(i)`

Again , `(x/a cos theta + y/b sin theta ) =1`

𝑆𝑞𝑢𝑎𝑟𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒, 𝑤𝑒 𝑔𝑒𝑡:

`(x/a cos theta + y/b sin theta )^2 = (1)^2`

`⇒ (x^2/a^2 cos^2 theta + y^2 /b^2 sin ^2 theta + 2 x/a xx y/b sin theta cos theta ) =     ....(ii)`

Now, adding (i) and (ii), we get:

`(x^2/a^2 sin^2 theta + y^2 /b^2 cos^2 theta -2 x/a xx y/b sin theta cos theta ) + (x^2/a^2 cos^2 theta + y^2 / b^2 sin^2 theta + 2 x/a xx y/b sin theta cos theta)`

 ⇒`x^2/a^2 sin^2 theta  + y^2/b^2 cos^2 theta + x^2 /a^2 cos^2 theta + y^2/b^2 sin^2 theta =2`

 ⇒`(x^2/a^2 sin^2 theta  + x^2/a^2 cos^2 theta)+(y^2/b^2 cos^2 theta + y^2/b^2 sin ^2 theta ) =2`

 ⇒`x^2/a^2 (sin^2 theta + cos^2 theta ) + y^2/b^2 (cos^2 theta + sin^2 theta ) =2`

 ⇒`x^2/a^2 + y^2 /b^2 =2     [∵ sin^2 theta + cos^2 theta =1]`

∴`x^2/a^2 + y^2/b^2 = 2`

We have x = a `cos^3 theta `

 = > `x/a = cos^3 theta     ........(i)`

 Again, `y = b  sin^3 theta`

  =  > `y/b = sin^3 theta      .....(ii)`

 Now, LHS = `(x/a)^(2/3) + (y/b)^(2/3)`

 = `( cos^3 theta )^(2/3) + (sin^3 theta )^ (2/3 )`     [ from (i) and (ii)]

 =` cos^2 theta + sin^2 theta `

 =1

𝐻𝑒𝑛𝑐𝑒, 𝐿𝐻𝑆 = 𝑅𝐻𝑆       

LHS = `(1 sin^2 theta + 2 sin theta + 1 + sin^2 theta -  2 sin theta)/(2 cos theta)`

`=> (2(1 + sin^2 theta))/(2 cos^2 theta) => (1 + sin^2 theta)/(1 - sin^2 theta)`     `[∵ cos^2 theta = 1 - sin^2 theta]`

∴ LHS = RHS Hence proved

We know `sin^2 A + cos^2 A = 1`

`sin^2 A = 1 - cos^2 A`

`=> sin^2 A . cosec^2 A`

`=> sin^2 A . 1/(sin^2 A) = 1`

∴ L.H.S = R.H.S

We know that `cosec^A - cot^2 A = 1`

So,

`(1 + cot^2 A)sin^2 A = cosec^2 A sin^2A`

`= (cosec A sin A)^2`

`= (1/sin A xx sin A)^2`

`= (1)^2`

= 1

We know that `sin^2 theta + cos^2 theta = 1`

Multiplying both numerator and the denominator by `(1 - sin theta)`, we have

`cos theta/(1 + sin theta) = (cos theta(1 - sin theta))/((1 + sin theta)(1 - sin theta))`

`= (cos theta(1 - sin theta))/(1 - sin^2 theta)`

`= (cos theta (1 - sin theta))/cos^2 theta`

`= (1 - sin theta)/cos theta`

L.H.S. = `cos^2 A + 1/(1 + cot^2 A)`

= `cos^2 A + 1/("cosec"^2 A)        ...[1 + cot^2A = "cosec"^2 A]`

= `cos^2 A + sin^2 A     ...[1/("cosec" A) = sin A]`

= `cos^2 A + sin^2 A`

= 1  (R.H.S.)       ...`[sin^2 A + cos^2A = 1]`

We have to prove  `(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`

We know that, `sin^2 theta + cos^2 theta = 1`

`(1 + sec theta)/sec theta = (1 + 1/cos theta)/(1/cos theta)`

`= ((cos theta + 1)/cos theta)/(1/cos theta)`

`= (1 + cos theta)/1`

Multiplying the numerator and denominator by `(1 - cos theta)` we have

`(1 + sec theta)/sec theta  = ((1 + cos theta)(1 - cos theta))/(1- cos theta)`

`= (1 - cos^2 theta)/(1- cos theta)`

`= sin^2 theta/(1 - cos theta)`

`sqrt((1 + sin A)/(1 - sin A)) = sec A + tan A`

LHS = `sqrt((1 + sin A)/(1 - sin A)`

Rationalize the numerator abd denominator with `sqrt(1 + sin A)`

LHS = `sqrt(((1 + sin A)(1 + sin A))/((1 - sin A)(1 + sin A)))`

= `sqrt((1 + sin A)^2/(1 - sin^2 A))`

= `sqrt((1 + sin A)^2/(cos^2 A))`

= `(1 + sin A)/(cos A)`

= `1/(cos A) + (sin A)/(cos A)`

= sec A + tan A

= RHS

In the given question, we need to prove `(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta)) = cot theta`

Using the property  `sin^2 theta + cot^2 theta = 1` we get

So

`(1 + cos theta - sin^2 theta)/(sin theta (1 +  cos theta))`

`= (1 + cos theta - (1 - cos^2 theta))/(sin theta (1 + cos theta)`

`= (cos theta + cos^2 theta)/(sin theta (1 + cos theta))`

Solving further, we get

`(cos theta + cos^2 theta)/(sin(1 + cos theta)) = (cos theta (1 + cos theta))/(sin theta(1 + cos theta))`

`= cos theta/sin theta`

`= cot theta`

Hence proved.

In the given question, we need to prove tan² A + cot² A = sec² A cosec² A − 2

Now using `tan theta = sin theta/cos theta` and `cot theta = cos theta/sin theta` in LHS we get

`tan^2 A + cot^2  A = sin^2 A/cos^2 A + cos^2 A/sin^2 A`

`= (sin^4 A + cos^4 A)/(cos^2 A sin^2 A)`

`= ((sin^2 A)^2 + (cos^2 A)^2)/(cos^2 A sin^2 A)`

Further, using the identity `a^2 + b^2 = (a + b)^2 - 2ab` we get

`((sin^2 A)^2 + (cos^2 A)^2)/(cos^2 A sin^2 A) = ((sin^2 A + cos^ A)^2 - 2 sin^2 A cos^2 A)/(sin^2 A cos^2 A)`

`= ((1)^2 - 2sin^2 A cos^2 A)/(sin^2 A cos^2 A)`

`= 1/(sin^2 A cos^2 A) - (2 sin^2 A cos^2 A)/(sin^2 A cos^2 A`

`= cosec^2 A sec^2 A - 2`

Since L.H.S = R.H.S

Hence proved.

In the given question, we need to prove `(cos theta)/(cosec theta + 1) + (cos theta)/(cosec theta - 1) = 2 tan theta`

Using the identity `a^2 - b^2  = (a + b)(a - b)`

`cos theta/((cosec theta + 1)) + cos theta/(cosec theta - 1) = (cos theta(cosec theta - 1)+ cos theta(cosec theta + 1))/(cosec^2 theta - 1)`

`= (cos theta (cosec theta - 1 + cosec theta + 1))/(cosec^2 theta -1)  = (cos theta(2 cosec theta))/cot^2 theta`

`= ((2 cos theta)(1/sin theta))/((cos^2 theta/sin^2 theta))`

`= 2 ((cos theta)/(sin theta))(sin^2 theta/cos^2 theta)`

`= 2 sin theta/cos theta`

`= 2 tan theta`

Hence proved.

∵ tan A = cot B

tan A = tan (90° – B)

A = 90° – B

A + B = 90°. Proved

R.H.S. = `(sec^2A)/("cosec"^2A)`

= `(1 + tan^2A)/(1 + cot^2A)`   ...`[(∵ 1 + tan^2A = sec^2A),(1 + cot^2A = "cosec"^2A)]`

= `(1 + (sin^2A)/(cos^2A))/(1 + (cos^2A)/(sin^2A))`

= `((cos^2A  +  sin^2A)/(cos^2A))/((sin^2A  +  cos^2A)/(sin^2A))`

= `(1/(cos^2A))/(1/(sin^2A))`   ...[∵ sin²A + cos²A = 1]

= `(sin^2A)/(cos^2A)`

= tan²A

= tan A . tan A

= `(tan A)/(cot A)`

= L.H.S.

∴ `(tan A)/(cot A) = (sec^2A)/("cosec"^2A)`

LHS = sec 70° sin 70° + cos 20° cosec 70°

= sec (90° - 20°) sin 20° + cos 20° cosec (90° - 20°) 

`= "cosec" 20°. 1/("cosec" 20°)+ 1/(sec 20°)  sec 20°`

= 1 + 1

= 2 

= RHS 

We have to prove `((1 + cot^2 theta) tan theta)/sec^2 theta = cot theta`

We know that, `sec^2 theta - tan^2 theta = 1`

So

`((1 + cot^2 theta)tan theta)/sec^2 theta = ((1 + cot^2 theta)tan theta)/(1 + tan^2 theta)`

` = ((1 + 1/tan^2 theta)tan theta)/(1 + tan^2 theta)`

`= (((tan^2 theta + 1)/(tan^2 theta)) tan theta)/(1 + tan^ 2 theta)`

`= ((1 + tan^2 theta)tan theta)/(tan^2 theta(1 + tan^2 theta))`

`= 1/tan theta`

`= cot theta`

We have to prove`(sec theta + cos theta)(sec theta - cos theta) = tan^2 theta + sin^2 theta`

We know that

`sin^2 theta + cos^2 theta = 1`

`sec^2 theta - tan^2 theta = 1`

`(sec theta + cos theta)(sec theta - cos theta) = sec^2 theta - cos^2 theta`

`= (1 + tan^2 theta) - (1 - sin^2 theta)`

`= 1 + tan^2 theta - 1 + sin^2 theta`

`= tan^2 theta + sin^2 theta`

For an acute angle A in a right-angled triangle:

• Hypotenuse is the side opposite the right angle.

• Perpendicular is the side opposite angle A.

• Base is the side adjacent to angle A.

sin² A + cos² A = 1

1 + tan² A = sec² A 

1 + cot² A = cosec² A

A trigonometric table consists of three parts:

1. A column on the extreme left containing degrees from 0^∘ to 89^∘

2. Ten columns headed by 0′, 6′, 12′, 18′, 24′, 30′, 36′, 42′, 48′ and 54′.

3. Five columns of mean differences headed by 1′, 2′, 3′, 4′ and 5′

4. Relation Between Degrees and Minutes
   1^∘ = 60′

5. Mean difference is added in case of:
   sine
   tangent
   secant

6. Mean difference is subtracted in the case of:
   cosine
   cotangent
   cosecant