Question

prove the left hand side of question ​

Answer 1

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

Applying the formula $a^3-b^3 = (a-b)(a^2+ab+b^2)$, we get

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

        $= tan^2\theta+tan\theta+1$

Applying the identity $Tan^2\theta+1 = Sec^2\theta$, we get

$LHS = Sec^2\theta+Tan\theta = RHS$

HOPE THIS HELPS!!