Prove the following identity,where the angles involved are acute angles for which the expressions are defined:
$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$

(A) $L.H.S. = \frac{\tan \theta}{1-\cot \theta} + \frac{\cot \theta}{1-\tan \theta}$
$= \frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$
$= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}$
$= \frac{\sin^2 \theta}{\cos \theta(\sin \theta-\cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta-\cos \theta)}$
$= \frac{1}{(\sin \theta-\cos \theta)} \left[ \frac{\sin^2 \theta}{\cos \theta} - \frac{\cos^2 \theta}{\sin \theta} \right]$
$= \left( \frac{1}{\sin \theta-\cos \theta} \right) \left[ \frac{\sin^3 \theta-\cos^3 \theta}{\sin \theta \cos \theta} \right]$
$= \left( \frac{1}{\sin \theta-\cos \theta} \right) \left[ \frac{(\sin \theta-\cos \theta)(\sin^2 \theta+\cos^2 \theta+\sin \theta \cos \theta)}{\sin \theta \cos \theta} \right]$
$= \frac{1+\sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \operatorname{cosec} \theta \sec \theta + 1 = R.H.S.$