Question

Prove:

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

Answer 1

$\frac{{tan}^3\theta -1}{tan\theta -1}=\frac{(tan\theta -1)({tan}^2\theta +tan\theta \ast 1+1)}{(tan\theta -1)}$

$[a^3-b^3=(a-b)(a^2+ab+b^2)]$

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