Inverse Trigonometric Functions — Class 12 Maths Solution

exemplar objective MCQ NCERT,Ex.2.3,Q.34,Page.39
Question

If $|x| \le 1$, then $2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$ is equal to

  • (a) $4{\tan ^{ - 1}}x$ ✓ Correct
  • (b) 0
  • (c) $\frac{\pi }{2}$
  • (d) $\pi$
Step-by-step Solution
Correct answer: option (a)

We have, $2{\tan ^{ - 1}}x + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}}$

Let $x = \tan \theta$
$therefore, 2{\tan ^{ - 1}}\tan \theta + {\sin ^{ - 1}}\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$

$= 2\theta + {\sin ^{ - 1}}\sin 2\theta$
$= 2\theta + 2\theta$
$= 4\theta$
$= 4{\tan ^{ - 1}}x$

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NCERT & Exemplar solution for CBSE Class 12 Mathematics, Inverse Trigonometric Functions. Curated by Sachin Sharma. Free for all students.