Question

Prove that $se{c}^6\theta -ta{n}^6\theta -3se{c}^2\theta ta{n}^2\theta$ = 1.

Answer 1

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

Taking LHS

${\sec }^6\theta -{\tan }^6\theta -3{\sec }^2\theta {\tan }^2\theta$

$=({\sec }^2\theta {)}^3-({\tan }^2\theta {)}^3-3{\sec }^2\theta {\tan }^2\theta$ $a^3-b^3=(a-b)(a^2+b^2+ab)$

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

$=1({\sec }^4\theta +{\tan }^4\theta -2{\sec }^2\theta {\tan }^2\theta )$

$({\sec }^2\theta -{\tan }^2\theta {)}^2$ or $({\tan }^2\theta -{\sec }^2\theta {)}^2$

$(1{)}^2$ or $(-1{)}^2$

$=1$