Question

Evaluate:

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

Answer 1

Now,

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

$=1+{\tan }^2\left({\tan }^{-1}2\right)+1+{\cot }^2\left({\cot }^{-1}3\right)$ [ Since ${\sec }^2\theta =1+ta{n}^2\theta$ and $cose{c}^2\theta =1+{\cot }^2\theta$]

$=1+\left(\tan \right({\tan }^{-1}2){)}^2+1+(\cot \left({\cot }^{-1}3\right){)}^2$

$=1+2^2+1+3^2$

$=15$.