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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
$\cos \left[ {{{\tan }^{ - 1}}\frac{1}{3} + {{\tan }^{ - 1}}\frac{1}{2}} \right] = $
  • ✓
    $\frac{1}{{\sqrt 2 }}$
  • B
    $\frac{{\sqrt 3 }}{2}$
  • C
    $\frac{1}{2}$
  • D
    $\frac{\pi }{4}$
✓

Answer

Correct option: A.
$\frac{1}{{\sqrt 2 }}$
a
(a) $\cos \,\left[ {{{\tan }^{ - 1}}\frac{1}{3} + {{\tan }^{ - 1}}\frac{1}{2}} \right] = \cos \,\left[ {{{\tan }^{ - 1}}\left( {\frac{{\frac{1}{3} + \frac{1}{2}}}{{1 - \frac{1}{3} \times \frac{1}{2}}}} \right)} \right]$

$ = \cos \,\{ {\tan ^{ - 1}}(1)\} = \cos \frac{\pi }{4} = \frac{1}{{\sqrt 2 }}$.

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