Question

Prove that:

$\frac{\tan \theta }{\left(1-\cot \theta \right)}+\frac{\cot \theta }{\left(1-\tan \theta \right)}=\left(1+\sec \theta \cos ec\theta \right)$

Answer 1

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

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

$=\frac{\tan \theta +\cot \theta -{\sec }^2\theta -{\csc }^2\theta +2}{2-(\cot \theta +\tan \theta )}$

${\sec }^2\theta +{\csc }^2\theta ={\sec }^2\theta {\csc }^2\theta$

$\tan \theta +\cot \theta =\sec \theta \csc \theta$

$=\frac{\sec \theta \csc \theta -{\sec }^2\theta {\csc }^2\theta +2}{2-\sec \theta \csc \theta }$

$=(2-\sec \theta \csc \theta )(1+\sec \theta \csc \theta )(2-\sec \theta csc\theta )$

$\therefore \frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=(1+\sec \theta \csc \theta )$.