Question
Prove the following identities:
$\frac{1-\tan^2\theta}{1+\tan^2\theta}=\big(\cos^2\theta-\sin^2\theta\big)$

Answer

$\text{LHS}=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
$=\frac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}$
$=\frac{\Big(\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}\Big)}{\Big(\frac{\cos^2\theta+\sin^\theta}{\cos^2\theta}\Big)}=\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}$
$=\frac{\big(\cos^2\theta-\sin^2\theta\big)}{1}=\big(\cos^2\theta-\sin^2\theta\big)$
$=\text{R.H.S.}$
$\therefore\text{R.H.S.}=\text{L.H.S.}$

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