Question

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

Answer 1

L.H.S$=\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }$

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

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

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

$=\frac{1}{(\sin \theta -\cos \theta )}\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta \cos \theta }$

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

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

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

$=\sec \theta \csc \theta +1=$R.H.S

Hence proved.