Question

If $\frac{\tan \theta }{1-\cos \theta }+\frac{\cot \theta }{1-\tan \theta }=m+\sec \theta \text{cosec}\theta$.Find $m$

Answer 1

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

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

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

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

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

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

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

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

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

$1+\sec \theta \csc \theta$

$\therefore m=1$