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

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.

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. = `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

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.

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.

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 = `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

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`

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

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.

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

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.

∵ tan A = cot B

tan A = tan (90° – B)

A = 90° – B

A + B = 90°. Proved

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