Question

Prove the following trigonometric identities.

`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`

Answer 1

We need to prove  `1/(secA - 1) + 1/(sec A + 1) = 2 cosec A cot A`

Solving the L.H.S, we get

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

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

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

So

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

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

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

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

= 2cosec A cot A

Answer 2

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

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

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

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

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

= 2 cosec A. cot A
= RHS
Hence proved.