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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
$\tan \left[ {{{\cos }^{ - 1}}\frac{4}{5} + {{\tan }^{ - 1}}\frac{2}{3}} \right] =$
  • A
    $6/17$
  • ✓
    $17/6$
  • C
    $7/16$
  • D
    $16/7$
✓

Answer

Correct option: B.
$17/6$
b
(b) $\tan \,\left[ {{{\cos }^{ - 1}}\frac{4}{5} + {{\tan }^{ - 1}}\frac{2}{3}} \right]$

$ = \tan \,\left[ {{{\tan }^{ - 1}}\frac{{\sqrt {\left( {1 - \frac{{16}}{{25}}} \right)} }}{{\frac{4}{5}}} + {{\tan }^{ - 1}}\frac{2}{3}} \right]$

$ = \tan \,\left[ {{{\tan }^{ - 1}}\left( {\frac{{\frac{3}{4} + \frac{2}{3}}}{{1 - \frac{3}{4}.\frac{2}{3}}}} \right)} \right] = \tan \,.\,{\tan ^{ - 1}}\frac{{17}}{6} = \frac{{17}}{6}$.

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