Question

Prove that $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta .co\sec \theta$

Answer 1

$L.H.S:$ Converting into $\sin \theta$ and $\cos \theta$ form,

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

$=\frac{1}{\sin \theta -\cos \theta }(\frac{{\sin }^2\theta }{\cos \theta }-\frac{{\cos }^2\theta }{\sin \theta })=\frac{1}{\sin \theta -\cos \theta }\left(\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta \cos \theta }\right)$

$=\frac{1}{\sin \theta -\cos \theta }\left(\frac{(\sin \theta -\cos \theta )({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta )}{\sin \theta \cos \theta }\right)$ .... $(\because a^3-b^3=(a-b\left)\right(a^2+ab+b^2\left)\right)$

$=\frac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }=1+\frac{1}{\sin \theta \cos \theta }=1+\sec \theta \text{cosec}\theta =R.H.S$

Hence proved.