Question

Prove the following:

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

Answer 1

L.H.S:

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

Taking LCM of the denominators,

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

Since, (1 + sec A)(1 – sec A) = 1 – sec²A

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

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

= `(2 tan A  *sec A)/(sec^2 A - 1)`

Since,

sec²A – tan²A = 1

sec²A – 1 = tan²A

= `(2 tan A * sec A)/(tan^2 A)` 

Since, sec A = `(1/cosA)` and tan A = `(sinA/cosA)`

= `(2secA)/tanA = (2cosA)/(cosA sinA)`

= `2/sinA`

= 2 cosec A  ...`(∵ 1/sinA = "cosec" A)`

= R.H.S

Hence proved.