Inverse Trigonometric Functions — Class 12 Maths Solution

ncert misc SA NCERT Misc. , Q.10 , Page 52
Question

${\cot ^{ - 1}}\left( {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right) = \frac{x}{2},\;\;x \in \left( {0,\;\;\frac{\pi }{4}} \right)$

Step-by-step Solution

L.H.S.
$= {\cot ^{ - 1}}\left\{ {\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} \times \frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}} \right\}$

$= {\cot ^{ - 1}}\left\{ {\frac{{(1 + \sin x) + (1 - \sin x) + 2\sqrt {1 - {{\sin }^2}x} }}{{(1 + \sin x) - (1 - \sin x)}}} \right\}$
$= {\cot ^{ - 1}}\left\{ {\frac{{2(1 + \cos x)}}{{2\sin x}}} \right\} = {\cot ^{ - 1}}\left( {\frac{{1 + \cos x}}{{\sin x}}} \right)$

$= {\cot ^{ - 1}}\left\{ {\frac{{2{{\cos }^2}(x/2)}}{{2\sin (x/2)\cos (x/2)}}} \right\}$
$= {\cot ^{ - 1}}\left( {\cot \frac{x}{2}} \right) = \frac{x}{2} = R.H.S.$

NCERT & Exemplar solution for CBSE Class 12 Mathematics, Inverse Trigonometric Functions. Curated by Sachin Sharma. Free for all students.