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

Gujarat BoardEnglish MediumSTD 12 ScienceMaths2. inverse trigonometric function1 Mark
MCQ
A value of ${\tan ^{ - 1}}\,\,\left( {\sin \,\left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{2}{3}} } \right)} \right)} \right)$ is
  • A
    $\frac {\pi }{4}$
  • B
    $\frac {\pi }{2}$
  • C
    $\frac {\pi }{3}$
  • ✓
    $\frac {\pi }{6}$
✓

Answer

Correct option: D.
$\frac {\pi }{6}$
d
Consider ${\tan ^{ - 1}}\left[ {\sin \left( {{{\cos }^{ - 1}}\sqrt {\frac{2}{3}} } \right)} \right]$

Let ${\cos ^{ - 1}}\sqrt {\frac{2}{3}}  = \theta  \Rightarrow \cos \theta  = \sqrt {\frac{2}{3}} $

$ \Rightarrow \sin \theta  = \sqrt {1 - {{\cos }^2}\theta }  = \sqrt {1 - \frac{2}{3}}  = \sqrt {\frac{1}{3}} $

$\therefore {\tan ^{ - 1}}\left[ {\sin \left( {{{\cos }^{ - 1}}\sqrt {\frac{2}{3}} } \right)} \right] = {\tan ^{ - 1}}\left[ {\sin \theta } \right]$

      $ = {\tan ^{ - 1}}\left[ {\sqrt {\frac{1}{3}} } \right] = {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$

      $ = \frac{\pi }{6}$

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