Question

How do you verify `(cot theta - tan theta)/(cot theta + tan theta) = cos 2 theta`?

Answer 1

Taking LHS as follows

`LHS=\frac{\cot\theta-\tan\theta}{\cot\theta+\tan\theta}`

`=\frac{\cos\theta/\sin\theta-\sin\theta/\cos\theta}{\cos\theta/\sin\theta+\sin\theta/\cos\theta}`

`=\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}`

`=\frac{\cos^2\theta-(1-\cos^2\theta)}{1}`

`=\cos^2\theta-1+\cos^2\theta`

`=2\cos^2\theta-1`

`=\cos2\theta`

`=RHS`

proved.

Answer 2

Shown below

Explanation:

Let `t = tan(a/2)`

`sina = (2t)/(1+t^2)`

`cosa = (1-t^2)/(1+t^2)`

`=> tan a = (2t)/(1-t^2)`

`LHS:`

`= ( (1-t^2)/(2t) - (2t)/(1-t^2) ) / ((1-t^2)/(2t) + (2t)/(1-t^2))`

`= (1-t^2)/(1+t^2)`

`= cos a`

`= RHS`

`=> (cot(a/2) - tan(a/2) ) / (cot(a/2) + tan(a/2) ) = cos a`

let `theta = a/2 => 2theta = a`

Hence `(cot(theta) - tan(theta) ) / (cot(theta) + tan(theta) ) = cos 2theta`

This idea of using `t= tan(theta/2)` can be used for a great amount of problems similar to this one, and even for integrals like `int 1/(1+sintheta) d theta` where its called weirstrass substitution