Question

What is the value of ${\sec }^2\left({\tan }^{-1}\right(\frac{5}{11}\left)\right)$ ?

• A. 121/96
• B. 217/921
• C. 146/121
• D. 267/121

Answer 1

The correct option is C 146/121

We know that ${\sec }^{2}A=1+{\tan }^{2}A$

Let $A={\tan }^{-1}\left(\frac{5}{11}\right)$

$\Rightarrow \tan A=\tan {\tan }^{-1}\left(\frac{5}{11}\right)=\frac{5}{11}$

$\therefore {\sec }^2\left({\tan }^{-1}\left(\frac{5}{11}\right)\right)=1+{\tan }^2\left({\tan }^{-1}\left(\frac{5}{11}\right)\right)$

$=1+{\left(\tan \left({\tan }^{-1}\left(5/11\right)\right)\right)}^2$

$=1+{\left(\frac{5}{11}\right)}^2$

$=1+\frac{25}{121}$

$=\frac{146}{121}$