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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 ^{ - 1}}\frac{4}{5} + {\tan ^{ - 1}}\frac{3}{5} = $
  • ✓
    ${\tan ^{ - 1}}\frac{{27}}{{11}}$
  • B
    ${\sin ^{ - 1}}\frac{{11}}{{27}}$
  • C
    ${\cos ^{ - 1}}\frac{{11}}{{27}}$
  • D
    None of these
✓

Answer

Correct option: A.
${\tan ^{ - 1}}\frac{{27}}{{11}}$
a
(a) ${\cos ^{ - 1}}\frac{4}{5} + {\tan ^{ - 1}}\frac{3}{5} $

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

$\left[ {{\rm{Since}}\,\,{{\cos }^{ - 1}}x = {{\tan }^{ - 1}}\frac{{\sqrt {(1 - {x^2})} }}{x}} \right]$

$ = {\tan ^{ - 1}}\frac{3}{4} + {\tan ^{ - 1}}\frac{3}{5} $

$= {\tan ^{ - 1}}\,\left( {\frac{{\frac{3}{4} + \frac{3}{5}}}{{1 - \frac{3}{4}.\frac{3}{5}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{27}}{{11}}} \right)$.

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