Question

Prove : $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \cos ec\theta$

Answer 1

Consider the L.H.S.

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

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

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

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

$=\frac{{\sin }^3\theta -{\cos }^3\theta }{sin\theta \cos \theta (\sin \theta -\cos \theta )}$

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

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

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

$=1+\sec \theta \mathrm{cosec}\theta$

Henced, proved.