Inverse Trigonometric Functions — Class 12 Maths Solution

We have, $\sin \left[ {2{{\tan }^{ - 1}}(0.75)} \right] = \sin \left( {2{{\tan }^{ - 1}}\frac{3}{4}} \right)$

$= \sin \left( {{{\sin }^{ - 1}}\frac{{2 \cdot \frac{3}{4}}}{{1 + \frac{9}{{16}}}}} \right) = \sin \left[ {{{\sin }^{ - 1}}\frac{{3/2}}{{25/16}}} \right]$

$= \sin \left[ {{{\sin }^{ - 1}}\left( {\frac{{48}}{{50}}} \right)} \right] = \sin \left[ {{{\sin }^{ - 1}}\left( {\frac{{24}}{{25}}} \right)} \right] = \frac{{24}}{{25}} = 0.96$