Question

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

Answer 1

We have,

$\begin{array}{c}=\frac{\tan \theta \left(1-\cot \theta \right)-\cot \theta \left(1-\tan \theta \right)}{\left(1-\tan \theta \right)\left(1-\cot \theta \right)}\\ =\frac{\tan \theta -1-\cot \theta +1}{\left(1-\tan \theta \right)\left(1-\cot \theta \right)}\\ =\frac{\tan \theta -\cot \theta }{\left(1-\tan \theta \right)\left(1-\cot \theta \right)}\\ =\frac{\frac{\sin \theta }{\cos \theta }-\frac{\cos \theta }{\sin \theta }}{\left(1-\frac{\sin \theta }{\cos \theta }\right)\left(1-\frac{\cos \theta }{\sin \theta }\right)}\\ =\frac{{\sin }^2\theta -{\cos }^2\theta }{\left(\cos \theta -\sin \theta \right)\left(\sin \theta -\cos \theta \right)}\\ =\frac{\sin \theta +\cos \theta }{\cos \theta -\sin \theta }\end{array}$

$=R.H.S$

Proved.