STD 12 - 2. inverse trigonometric function — JEE कक्षा 11 साइन्स — Question

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

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

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

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