Revision: Trigonometry >> Introduction to Trigonometry Maths English Medium Class 10 CBSE

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

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

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

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.

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

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.

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