If $\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=\frac{5 \pi^2}{8}$, then the value of $x$ is

If $\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=\frac{5 \pi^2}{8}$, then the value of $x$ is
  1. $-2$
  2. $-1$
  3. 1
  4. 2

Solution

$\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=\frac{5 \pi^2}{8}$ $\Rightarrow\left(\tan ^{-1} x+\cot ^{-1} x\right)^2-2 \tan ^{-1} x \cot ^{-1} x=\frac{5 \pi^2}{8}$ $\Rightarrow\left(\tan ^{-1} x+\cot ^{-1} x\right)^2$ $-2 \tan ^{-1} x\left(\frac{\pi}{2}-\tan ^{-1} x\right)=\frac{5 \pi^2}{8}$ $\begin{aligned} & \Rightarrow \frac{\pi^2}{4}-2 \times \frac{\pi}{2} \tan ^{-1} x+2\left(\tan ^{-1} x\right)^2=\frac{5 \pi^2}{8} \\ & \Rightarrow 2\left(\tan ^{-1} x\right)^2-\pi \tan ^{-1} x-\frac{3 \pi^2}{8}=0 \\ & \Rightarrow \tan ^{-1} x=-\frac{\pi}{4}, \frac{3 \pi}{4} \\ & \Rightarrow \tan ^{-1} x=-\frac{\pi}{4} \Rightarrow x=-1\end{aligned}$

Asked in: MHT CET 2023 (13 May Shift 2)

Practice more Inverse Trigonometric Functions questions on Aicharya