Question

Prove that: $\frac{\sin \theta }{(1-\cot \theta )}+\frac{\cos \theta }{(1-\tan \theta )}=\cos \theta +\sin \theta$

Answer 1

$LHS=\frac{\sin \theta }{1-\cot \theta }+\frac{\cos \theta }{1-\tan \theta }$

$=\frac{\sin \theta }{1-\left(1/\tan \theta \right)}+\frac{\cos \theta }{1-\tan \theta }$

$=\frac{\sin \theta \tan \theta }{\tan \theta -1}+\frac{\cos \theta }{1-\tan \theta }$

$=\frac{\sin \theta \tan \theta -\cos \theta }{\tan \theta -1}$

$=\frac{\frac{{\sin }^2\theta }{\cos \theta }-\cos \theta }{\tan \theta -1}$

$=\frac{{\sin }^2\theta -{\cos }^2\theta }{\cos \theta (\tan \theta -1)}$

$=\frac{(\sin \theta +\cos \theta )(\sin \theta -\cos \theta )}{\cos \theta (\sin \theta /\cos \theta -1)}$

$=\frac{(\sin \theta +\cos \theta )(\sin \theta -\cos \theta )}{\sin \theta -\cos \theta )}$

$=\sin \theta +\cos \theta$