Question

Prove the following trigonometric identities

tan² A + cot² A = sec² A cosec² A − 2

Answer 1

In the given question, we need to prove tan² A + cot² A = sec² A cosec² A − 2

Now using `tan theta = sin theta/cos theta` and `cot theta = cos theta/sin theta` in LHS we get

`tan^2 A + cot^2  A = sin^2 A/cos^2 A + cos^2 A/sin^2 A`

`= (sin^4 A + cos^4 A)/(cos^2 A sin^2 A)`

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

Further, using the identity `a^2 + b^2 = (a + b)^2 - 2ab` we get

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

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

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

`= cosec^2 A sec^2 A - 2`

Since L.H.S = R.H.S

Hence proved.