Question

Solve: $\frac{tan\theta }{1-cot\theta }+\frac{cot\theta }{1-tan\theta }=1+sec\theta cosec\theta$

Answer 1

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

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

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

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

$=\frac{{\sin }^3\theta -{\cos }^3\theta }{\cos \theta \sin \theta (\sin \theta -\cos \theta )}$ Using $a^3-b^3(a-b)(a^2+ab+b^2)$

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

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

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

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

Hence proved