Question

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

Answer 1

Now,

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

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

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

$={\sec }^2\theta +\tan \theta$. [ Since ${\tan }^2\theta +1={\sec }^2\theta$]