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JEE Main 201219 MayOnline
Exam
JEE Main
Year
2012
Date
19 May
Mode
Online
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2012 · 19 May · Online · Mathematics · Inverse Trigonometric Functions

A value of ^ -1 ( ( ^ -1 ( (2)/(3) ))) is

A value of tan⁡−1\tan ^{-1}\left(\sin \left(\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)\right)\right) is

Options

  1. Aπ4\frac{\pi}{4}
  2. Bπ2\frac{\pi}{2}
  3. Cπ3\frac{\pi}{3}
  4. Dπ6\frac{\pi}{6}

Correct answer

D. π6\frac{\pi}{6}

Step-by-step solution

Consider \tan ^{-1}\left[\sin \left(\cos ^{-1} \sqrt{\frac{2}{3}}\right)\right]  Let cos⁡−123=θ⇒cos⁡θ=23⇒sin⁡θ=1−cos⁡2θ=1−23=13∴tan⁡−1[sin⁡(cos⁡−123)]=tan⁡−1[sin⁡θ]=tan⁡−1[13]=tan⁡−1(13)=π6\begin{aligned} & \text{ Let } \cos ^{-1} \sqrt{\frac{2}{3}}=\theta \Rightarrow \cos \theta=\sqrt{\frac{2}{3}} \\ & \Rightarrow \quad \sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{2}{3}}=\sqrt{\frac{1}{3}} \\ & \therefore \quad \tan ^{-1}\left[\sin \left(\cos ^{-1} \sqrt{\frac{2}{3}}\right)\right]=\tan ^{-1}[\sin \theta] \\ & =\tan ^{-1}\left[\sqrt{\frac{1}{3}}\right]=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ & =\frac{\pi}{6} \end{aligned}
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