Revision: Trigonometry JEE Main Trigonometry

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

The straight line joining the eye of the observer to the point on the object being viewed.

The angle between the line of sight and the horizontal through the observer’s eye, when the object is below the level of the observer’s eye.

The angle between the line of sight and the horizontal through the observer’s eye, when the object is above the level of the observer’s eye.

Two angles are said to be allied when their sum or difference is either 0 (zero) or an integral multiple of n/2.

Examples:

• −θ,  π2 ± θ,  π ± θ, 3π2 ± θ, 2π ± θ,etc.

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

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

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

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

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

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

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`

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 = `(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 `(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

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

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.

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

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 `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 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`

`(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.

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.

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       

Given that: m sinθ = n sin(θ + 2α)

⇒ `(sin(theta + 2alpha))/sintheta = m/n`

Using componendo and dividendo theorem, we get,

⇒ `(sin(theta + 2alpha) + sintheta)/(sin(theta + 2alpha) - sintheta) = (m + n)/(m - n)`

⇒ `(2sin((theta + 2alpha + theta)/2).cos((theta + 2alpha - theta)/2))/(2cos((theta + 2alpha + theta)/2).sin((theta + 2alpha - theta)/2)) = (m + n)/(m - n)`    .......`[(because sinA + sinB = 2sin  (A + B)/2 . cos  (A - B)/2),(sinA - sinB = 2cos  (A + B)/2 . sin  (A - B)/2)]`

⇒ `(sin(theta + alpha).cosalpha)/(cos(theta + alpha).sinalpha) = (m + n)/(m - n)`

⇒ `tan(theta + alpha)cotalpha = (m + n)/(m - n)`

Hence proved.

Given that: cosα + cosβ = 0

And sinα + sinβ = 0

∴ (cosα + cosβ)² - (sinα + sinβ)² = 0

⇒ (cos²α + cos²β + 2cosα cosβ) - (sin²α + sin²β + 2sinα sinβ) = 0

⇒ cos²α + cos²β + 2 cosα cosβ - sin²α - sin²β - 2 sinα sinβ = 0

⇒ (cos²α - sin²α) + (cos²β - sin²β) + 2(cosα cosβ - sinα sinβ) = 0

⇒ cos2α + cos2β + 2cos(α + β) = 0

Hence, cos2α + cos2β = -2cos(α + β).

Hence proved

Given that: cos(α + β) = `4/5`

∴ tan(α + β) = `3/4`

And sin(α – β) = `5/13`

∴ tan(α – β) = `5/12`

Now tan 2α = tan[α + β + α – β]

= tan[(α + β) + (α – β)]

= `(tan(alpha + beta) + tan(alpha - beta))/(1 - tan(alpha + beta).tan(alpha - beta))`

= `(3/4 + 5/12)/(1 - 3/4 xx 5/12)`

= `((9 + 5)/12)/((48 - 15)/48)`

= `14/12 xx 48/33`

= `56/33`

Hence, tan 2α = `56/33`.

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)`

sin² A + cos² A = 1

1 + tan² A = sec² A 

1 + cot² A = cosec² A

i. \[\sin^{-1}\frac{1}{x}=\mathrm{cosec}^{-1}x\] if x ≥ 1 or x ≤ −1
\[\cos^{-1}\frac{1}{x}=\sec^{-1}x\] if x ≥ 1 or x ≤ −1
\[\tan^{-1}\frac{1}{x}=\cot^{-1}x\] if x > 0

ii. sin⁻¹(−x) = −sin⁻¹x, for x ∈ [−1, 1]
tan⁻¹(−x) = −tan⁻¹x, for x ∈ R
cosec⁻¹(−x) = −cosec⁻¹x, for x ≥ 1
cos⁻¹(−x) = π − cos⁻¹x, for x ∈ [−1, 1]
sec⁻¹(−x) = π − sec⁻¹x, for x ≥ 1
cot⁻¹(−x) = π − cot⁻¹x, for x ∈ R

\[\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2},\] for x ∈ [−1, 1]

\[\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2},\] for x ∈ R

\[\sec^{-1}x+\cos\sec^{-1}x=\frac{\pi}{2},\] for |x| ≥ 1

\[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\frac{x+y}{1-xy}\right),\] for x > 0, y > 0 and xy < 1

\[\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right),\] for x, y > 0 and xy > 1

\[\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right),\] for x, y > 0

\[2\tan^{-1}x=\sin^{-1}\left(\frac{2x}{1+x^{2}}\right),\] if −1 ≤ x ≤ 1

\[2\tan^{-1}x=\cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right),\] if x > 0

\[2\tan^{-1}x=\tan^{-1}\left(\frac{2x}{1-x^{2}}\right),\] if −1 < x < 1

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.