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2. inverse trigonometric function — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths2. inverse trigonometric function1 Mark
MCQ
If ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + {\sin ^{ - 1}}\left( {\frac{{2b}}{{1 + {b^2}}}} \right) = 2{\tan ^{ - 1}}x,$ then $x = $
  • A
    $\frac{{a - b}}{{1 + ab}}$
  • B
    $\frac{b}{{1 + ab}}$
  • C
    $\frac{b}{{1 - ab}}$
  • ✓
    $\frac{{a + b}}{{1 - ab}}$
✓

Answer

Correct option: D.
$\frac{{a + b}}{{1 - ab}}$
d
(d) ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + {\sin ^{ - 1}}\left( {\frac{{2b}}{{1 + {b^2}}}} \right) = 2{\tan ^{ - 1}}x$

Putting $a = \tan \theta $ and $b = \tan \phi $

So, ${\sin ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) + {\sin ^{ - 1}}\left( {\frac{{2\tan \phi }}{{1 + {{\tan }^2}\phi }}} \right) = 2{\tan ^{ - 1}}x$

==> ${\sin ^{ - 1}}\sin (2\theta ) + {\sin ^{ - 1}}\sin (2\phi ) = 2{\tan ^{ - 1}}x$

==> $2(\theta + \phi ) = 2{\tan ^{ - 1}}x$

Hence $x = \tan (\theta + \phi )$

==> $x = \frac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }}$

Substituting these values, we get $x = \frac{{a + b}}{{1 - ab}}$.

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