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STD 12 - 2. inverse trigonometric function — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 2. inverse trigonometric function1 Mark
MCQ
If ${\tan ^{ - 1}}\frac{{a + x}}{a} + {\tan ^{ - 1}}\frac{{a - x}}{a} = \frac{\pi }{6}$,then ${x^2} =$
  • A
    $2\sqrt 3 a$
  • B
    $\sqrt 3 a$
  • ✓
    $2\sqrt 3 {a^2}$
  • D
    None of these
✓

Answer

Correct option: C.
$2\sqrt 3 {a^2}$
c
(c) Given equation is ${\tan ^{ - 1}}\frac{{a + x}}{a} + {\tan ^{ - 1}}\frac{{a - x}}{a} = \frac{\pi }{6}$

$ \Rightarrow \,{\tan ^{ - 1\,}}\left( {\frac{{\frac{{a + x}}{a} + \frac{{a - x}}{a}}}{{1 - \frac{{a + x}}{a}.\,\frac{{a - x}}{a}}}} \right) = \frac{\pi }{6}$

$ \Rightarrow \,\,\frac{{2{a^2}}}{{{x^2}}} = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}\,\, $

$\Rightarrow \,\,{x^2} = 2\sqrt 3 {a^2}$.

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