Question

Prove the following trigonometric identities

`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`

Answer 1

We have to prove `(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`

Consider the expression

L.H.S

`(1 + tan^2 theta)/(1 + cot^2 theta) = (1 + tan^2 theta)/(1 + 1/(tan^2 theta))`

= `(1 +tan^2 theta)/((tan^2 theta + 1)/tan^2 theta)`

`= tan^2 theta (1 + tan^2 theta)/(1 + tan^2 theta)`

`= tan^2 theta` 

= R.H.S

Again, we have 

L.H.S

`((1 - tan theta)/(1 - cot theta))^2 = ((1 - tan theta)/(1 - 1/(tan theta)))^2`

`= ((1 - tan theta)/((tan theta - 1)/tan theta))^2`

`=[(tantheta(1-tantheta))/-(1-tantheta)]^2`

`=(-tantheta)^2=tan^2theta`

= R.H.S