Question

${\sec }^2({\tan }^{-1}2)+cose{c}^2({\cot }^{-1}3)=$

• A. 5
• B. 10
• C. 15
• D. 20

Answer 1

Answer: (C)

$se{c}^2\left(ta{n}^{-1}2\right)=1+ta{n}^2\left(ta{n}^{-1}2\right)$
$=1+{\left(2\right)}^2=5$
$cose{c}^2\left(co{t}^{-1}3\right)=1+co{t}^2\left(co{t}^{-1}3\right)$
$=1+{\left(3\right)}^2=10$
$\therefore$ reqd. value $=5+10$
$=15$