Prove that: tan²(sec⁻¹ 2) + cot²(cosec⁻¹ 3) = 11.

Prove that: $\tan^2(\sec^{-1} 2) + \cot^2(\operatorname{cosec}^{-1} 3) = 11$.

Given: $\tan^2(\sec^{-1} 2) + \cot^2(\operatorname{cosec}^{-1} 3)$

To show: $11$

Answer

Hence $\tan^2(\sec^{-1}2)+\cot^2(\operatorname{cosec}^{-1}3)=11$.

Let $\sec^{-1}2=\theta$, so $\sec\theta=2$; using $\tan^2\theta=\sec^2\theta-1$: $\tan^2(\sec^{-1}2)=2^2-1=3$.
Let $\operatorname{cosec}^{-1}3=\phi$, so $\operatorname{cosec}\phi=3$; using $\cot^2\phi=\operatorname{cosec}^2\phi-1$: $\cot^2(\operatorname{cosec}^{-1}3)=3^2-1=8$.
Adding: $\tan^2(\sec^{-1}2)+\cot^2(\operatorname{cosec}^{-1}3)=3+8$.
$=11$, which is the required result.