Inverse Trigonometric Functions — Class 12 Maths Solution

ncert misc SA NCERT Misc. , Q.4 , Page 51
Question

${\sin ^{ - 1}}\frac{8}{{17}} + {\sin ^{ - 1}}\frac{3}{5} = {\tan ^{ - 1}}\frac{{77}}{{36}}$

Step-by-step Solution

Let ${\sin ^{ - 1}}\frac{8}{{17}} = x \Rightarrow \sin x = \frac{8}{{17}} \Rightarrow \tan x - \frac{8}{{15}}$
Let ${\sin ^{ - 1}}\frac{3}{5} = y \Rightarrow \frac{3}{5} = \sin y \Rightarrow \tan y = \frac{3}{4} \Rightarrow y = {\tan ^{ - 1}}\frac{3}{4}$

$\Rightarrow$ $x + y = {\tan ^{ - 1}}\frac{8}{{15}} + {\tan ^{ - 1}}\frac{3}{4}$

$= {\tan ^{ - 1}}\left( {\frac{{\frac{8}{{15}} + \frac{3}{4}}}{{1 - \frac{8}{{15}} \times \frac{3}{4}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{\frac{{32 + 45}}{{60}}}}{{1 - \frac{{24}}{{60}}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{77}}{{36}}} \right)$

← All Inverse Trigonometric Functions solutions Practice tests →

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