Question

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

Answer 1

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

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

$=\frac{\tan \theta }{{\displaystyle \frac{\tan \theta -1}{\tan \theta }}}+\frac{{\displaystyle 1}}{\tan \theta }\times \frac{1}{1-\tan \theta }$

$=\frac{{\tan }^2\theta }{{\displaystyle \tan \theta -1}}+\frac{1}{\tan \theta (1-\tan \theta )}$

$=\frac{{\tan }^2\theta }{{\displaystyle \tan \theta -1}}-\frac{1}{\tan \theta (\tan \theta -1)}$

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

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

$=\frac{{\tan }^2\theta +\tan \theta +1}{\tan \theta }$

$=\tan \theta +1+cot\theta$

$=1+\tan \theta +cot\theta$

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

$=1+\frac{si{n}^2\theta +{\cos }^2\theta }{\cos \theta \sin \theta }$

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

$=1+sec\theta \cos ec\theta$

$\therefore L.H.S=R.H.S.$

Hence proved