Question

Prove that :$\frac{{\tan }^3\theta -1}{\tan \theta -1}={\sec }^2\theta +\tan \theta$

Answer 1

$\frac{1+{\tan }^3\theta }{1+\tan \theta }+\tan \theta -{\sec }^2\theta$

${\sec }^2\theta -{\tan }^2\theta =1$

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

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

$1+{\tan }^2\theta -\tan \theta +\tan \theta -1-{\tan }^2\theta$

$=0$

$\therefore \frac{1+{\tan }^3\theta }{1+\tan \theta }+\tan \theta -{\sec }^2\theta =0$