Question

If ${\sec }^2\alpha +{\tan }^2\alpha =\frac{13}{12}$ then ${\sec }^4\alpha -{\tan }^4\alpha =...$

• A. $1$
• B. $-1$
• C. $0$
• D. $\frac{13}{12}$

Answer 1

The correct option is D $\frac{13}{12}$

${\sec }^2\alpha +{\tan }^2\alpha =\frac{13}{12}$

${\sec }^2\alpha =1+{\tan }^2\alpha$

$1+2{\tan }^2\alpha =\frac{13}{12}$

$2{\tan }^2\alpha =\frac{13}{12}-1=\frac{1}{12}$

${\tan }^2\alpha =\frac{1}{24}$

${\tan }^4\alpha =\frac{1}{{24}^2}=\frac{1}{576}$

${\sec }^2\alpha =1+{\tan }^2\alpha =\frac{25}{24}$

${\sec }^4\alpha =\frac{625}{576}$

${\sec }^4\alpha -{\tan }^4\alpha =\frac{625-1}{576}=\frac{624}{576}=\frac{13}{12}$