Inverse Trigonometric Functions — Class 12 Maths Solution

We have, $\cot \left[ {{{\cos }^{ - 1}}\left( {\frac{7}{{25}}} \right)} \right]$

Let ${\cos ^{ - 1}}\frac{7}{{25}} = x$
$\Rightarrow$ $\cos x = \frac{7}{{25}}$
$therefore, \sin x = \sqrt {1 - {{\cos }^2}x} = \sqrt {1 - {{\left( {\frac{7}{{25}}} \right)}^2}}$

$= \sqrt {\frac{{625 - 49}}{{625}}} = \frac{{24}}{{25}}$

$therefore, \cot x = \frac{{\cos x}}{{\sin x}} = \frac{{\frac{7}{{25}}}}{{\frac{{24}}{{25}}}} = \frac{7}{{24}}$

……(i)
$\Rightarrow$ $x = {\cot ^{ - 1}}\left( {\frac{7}{{24}}} \right) = {\cos ^{ - 1}}\left( {\frac{7}{{25}}} \right)$

$therefore, \cot \left( {{{\cos }^{ - 1}}\frac{7}{{25}}} \right) = \cot \left( {{{\cot }^{ - 1}}\frac{7}{{24}}} \right) = \frac{7}{{24}}$