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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

If ^ -1 ((1-x)/(1+x))-(1)/(2) ^ -1 x=0 , for x>0 , then x=

If tan⁡−1(1−x1+x)−12tan⁡−1x=0\tan ^{-1}\left(\frac{1-x}{1+x}\right)-\frac{1}{2} \tan ^{-1} x=0, for x>0x>0, then x=x=

Options

  1. A3\sqrt{3}
  2. B12\frac{1}{\sqrt{2}}
  3. C13\frac{1}{\sqrt{3}}
  4. D13\frac{1}{3}

Correct answer

C. 13\frac{1}{\sqrt{3}}

Step-by-step solution

Here tan⁡−1(1−x1+x)=12tan⁡−1x⇒tan⁡−1[1−x1+(1)(x)]=12tan⁡−1x∴tan⁡−1(1)−tan⁡−1x=12tan⁡−1xπ4=32tan⁡−1x⇒tan⁡−1x=π4×23⇒tan⁡−1x=π6∴x=tan⁡π6=13\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x \Rightarrow \tan ^{-1}\left[\frac{1-x}{1+(1)(x)}\right]=\frac{1}{2} \tan ^{-1} x\therefore \tan ^{-1}(1)-\tan ^{-1} x=\frac{1}{2} \tan ^{-1} x\frac{\pi}{4}=\frac{3}{2} \tan ^{-1} x \Rightarrow \tan ^{-1} x=\frac{\pi}{4} \times \frac{2}{3} \Rightarrow \tan ^{-1} x=\frac{\pi}{6}\therefore x=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}
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