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

2 Answers
Harish Chandra Rajpoot
Jun 29, 2018

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.

Rhys
Jun 30, 2018

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

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