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

Answer

$\text{L.H.S.}=\frac{\cos^2\theta}{(1-\tan\theta)}+\frac{\sin^2\theta}{(\sin\theta-\sin\theta)}$ $=\frac{\cos^2\theta}{\Big(1-\frac{\sin\theta}{\cos\theta}\Big)}+\frac{\sin^3\theta}{(\sin\theta-\cos\theta)}$ $=\frac{\cos^3\theta}{(\cos\theta-\sin\theta)}+\frac{\sin^3\theta}{(\sin\theta-\cos\theta)}$$=\frac{\cos^3\theta}{(\cos\theta-\sin\theta)}-\frac{\sin^3\theta}{\cos\theta-\sin\theta}$
$=\frac{\cos^3\theta-\sin^3\theta}{(\cos\theta-\sin\theta)}$ $=\frac{(\cos\theta-\sin\theta)\big(\cos^2\theta+\cos\theta\sin\theta+\sin^2\theta\big)}{(\cos\theta-\sin\theta)}$ $\Big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})\big(\text{a}^2+\text{ab}+\text{b}^2\big)\Big]$ $=(1+\cos\theta\sin\theta)$ $=\text{R.H.S.}$ $\therefore\text{R.H.S.}=\text{L.H.S.}$

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