Question

Prove that ${\tan }^2\left({\sec }^{-1}2\right)+{\cot }^2\left(cose{c}^{-1}3\right)=11$.

Answer 1

LHS

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

$={\sec }^2\left({\sec }^{-1}2\right)-1+co{\sec }^2\left(co{\sec }^{-1}3\right)-1[{\sec }^2\theta -{\tan }^2\theta =1;co{\sec }^2\theta -{\cot }^2\theta =1]$

$={\left[\sec \left({\sec }^{-1}2\right)\right]}^2-1+{\left[co\sec \left(co{\sec }^{-1}3\right)\right]}^2-1=2^2-1+4^2-1=4-1+9-1=3+8=11=$ RHS