Question

The value of ${\sec }^2({\tan }^{-1}3)+cose{c}^2({\cot }^{-1}2)$ is equal to

• A. 5
• B. 13
• C. 15
• D. 23

Answer 1

Answer: (C)

We have $se{c}^2\left(ta{n}^{-1}3\right)+case{c}^2\left(co{t}^{-1}2\right)$
$={\left(sec\left(ta{n}^{-1}3\right)\right)}^2+{\left(cosec\left(co{t}^{-1}2\right)\right)}^2$
$={\left(sec\left(se{c}^{-1}\sqrt{10}\right)\right)}^2+{\left(cosec\left(cose{c}^{-1}\sqrt{5}\right)\right)}^2$
$=\left(\sqrt{10}\right)+{\left(\sqrt{5}\right)}^2=15$