Question

Prove ${\tan }^6\theta +3{\tan }^4\theta +3{\tan }^2={\sec }^6\theta -1$

Answer 1

$LHS={\tan }^6\theta +3{\tan }^4\theta +3{\tan }^2\theta ={\tan }^6\theta +3{\tan }^2\theta (1+{\tan }^2\theta )={\tan }^6\theta +3{\tan }^2\theta {\text{sec}}^2\theta$

$=({\text{sec}}^2\theta -1{)}^3+3({\text{sec}}^2-1){\text{sec}}^2\theta$

$={\text{sec}}^6\theta -3{\text{sec}}^4\theta +3{\text{sec}}^2\theta -1+3{\text{sec}}^4\theta -3{\text{sec}}^2\theta ={\text{sec}}^6\theta -1$

$LHS=RHS$

hence proved