Inverse Trigonometric Functions — Class 12 Maths Solution

ncert misc SA NCERT Misc. , Q.9 , Page 52
Question

${\tan ^{ - 1}}\sqrt x = \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right),\;\;x \in [0,\;1]$

Step-by-step Solution

Putting $x = {\tan ^2}\theta ,$ we get
R.H.S.$= \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right) = \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$= \frac{1}{2}{\cos ^{ - 1}}(\cos 2\theta ) = \frac{1}{2} \times 2\theta = \theta = {\tan ^{ - 1}}\sqrt x = R.H.S.$

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

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