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Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$
Solution
$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$
$L.H.S.$ $=\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$
$=\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$
$=\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}$
$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}-\frac{\cos ^{2} \theta}{\sin \theta(\sin \theta-\cos \theta)}$
$=\frac{1}{(\sin \theta-\cos \theta)}\left[\frac{\sin ^{2} \theta}{\cos \theta}-\frac{\cos ^{2} \theta}{\sin \theta}\right]$
$=\left(\frac{1}{\sin \theta-\cos \theta}\right)\left[\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta \cos \theta}\right]$
$=\left(\frac{1}{\sin \theta-\cos \theta}\right)\left[\frac{(\sin \theta-\cos \theta)\left(\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta\right)}{\sin \theta \cos \theta}\right]$
$=\frac{(1+\sin \theta \cos \theta)}{(\sin \theta \cos \theta)}$
$=\sec \theta \operatorname{cosec} \theta+1$
$= R . H.S.$
