Question

Solve it:-
${\sec }^{6}x-{\tan }^{6}x=1+3{\sec }^{2}x\times {\tan }^{2}x=1$

Answer 1

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

Taking LHS

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

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

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

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

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

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

1