Question

Prove the following.

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

Answer 1

\[\frac{\tan^3 \theta - 1}{\tan\theta - 1}\]

\[ = \frac{\left( \tan\theta - 1 \right)\left( \tan^2 \theta + \tan\theta \times 1 + 1 \right)}{\tan\theta - 1} \left[ a^3 - b^3 = \left( a - b \right)\left( a^2 + ab + b^2 \right) \right]\]

\[ = \tan^2 \theta + \tan\theta + 1\]   

\[ = \sec^2 \theta + \tan\theta \left( 1 + \tan^2 \theta = \sec^2 \theta \right)\]