Question

Prove the following trigonometric identities.

`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`

Answer 1

We need to prove `(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`

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

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

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

Now using `sec theta = 1/cos theta` and `tan theta = sin theta/cos theta` we get

`sec^2 A + ((sec^2 A)/(tan^2 A)) = 1/cos^2 A + ((1/cos^2 A)/((sin^2 A)/(cos^2 A)))`

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

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

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

Further, using the property, `sin^2 theta + cos^2 theta = 1` we get

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

`= 1/((1 - sin^2 A)(sin^2 A))`     (using `cos^2 theta = 1 - sin^2 theta`)

`= 1/(sin^2 A - sin^4 A)`

Hence proved