Question

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

Answer 1

Given,

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

LHS:

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

using $a^3-b^3$ formula, we get,

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

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

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

:RHS