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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

( ^ -1 x)^2+( ^ -1 x)^2=(5 π^2)/(8) x=

\left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=\frac{5 \pi^2}{8} \Rightarrow x=

Options

  1. A-1
  2. B1
  3. C0
  4. Dπ58\pi \sqrt{\frac{5}{8}}

Correct answer

A. -1

Step-by-step solution

Given equation is (tan⁡−1x)2+(cot⁡−1x)2=5π28(tan⁡−1x+cot⁡−1x)2−2tan⁡−1x⋅cot⁡−1x=5x28[∵tan⁡−1x+cot⁡−1x=π2]\begin{gathered} \left(\tan ^{-1} x\right)^2+\left(\cot ^{-1} x\right)^2=\frac{5 \pi^2}{8} \\ \left(\tan ^{-1} x+\cot ^{-1} x\right)^2-2 \tan ^{-1} x \cdot \cot ^{-1} x=\frac{5 x^2}{8} \\ {\left[\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right]} \end{gathered}⇒π24−2tan⁡−1x(π/2−tan⁡−1x)=5π28⇒−2⋅πtan⁡−1x2+2(tan⁡−1x)2=5π28−π24⇒(tan⁡−1x)2−πtan⁡−1x2=3π216\begin{aligned} & \Rightarrow \quad \frac{\pi^2}{4}-2 \tan ^{-1} x\left(\pi / 2-\tan ^{-1} x\right)=\frac{5 \pi^2}{8} \\ & \Rightarrow-2 \cdot \frac{\pi \tan ^{-1} x}{2}+2\left(\tan ^{-1} x\right)^2=\frac{5 \pi^2}{8}-\frac{\pi^2}{4} \\ & \Rightarrow \quad\left(\tan ^{-1} x\right)^2-\frac{\pi \tan ^{-1} x}{2}=\frac{3 \pi^2}{16}\end{aligned}$\begin{array}{cc}\therefore \quad & \tan ^{-1} x=\frac{\frac{\pi}{2} \pm \sqrt{\frac{\pi^2}{4}+\frac{3 \pi^2}{4}}}{2} \\ \Rightarrow \quad & \tan ^{-1} x=\frac{\frac{\pi}{2} \pm \pi}{2} \\ & =\frac{3 \pi}{4},-\frac{\pi}{4}\end{array}⇒x=tan⁡(3π4),tan⁡(−π4)⇒x=−1,−1\begin{array}{ll}\Rightarrow& x=\tan \left(\frac{3 \pi}{4}\right), \tan \left(-\frac{\pi}{4}\right)\\ \Rightarrow & x=-1,-1\end{array}$
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