Question

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

• A. $\frac{12}{13}$
• B. $\frac{13}{12}$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$\frac{13}{12}$

Acording to question,

$se{c}^4\theta -ta{n}^4\theta =(se{c}^2\theta -ta{n}^2\theta )(se{c}^2\theta +ta{n}^2\theta )$

=> $1\times (se{c}^2\theta +ta{n}^2\theta )$ ...........since, $(se{c}^2\theta -ta{n}^2\theta =1)$

=> $\frac{13}{12}$