8 - 4 - 1 3 = 4 - 1 3

8 - 4 - 1 3 = 4 = 4 - 1 3 = 4 - 1 ( 3 ) Hence, L.H.S. = R.H.S. Solve the following equations : ;

Inverse Trigonometric Functions — Class 12 Maths Solution

ncert misc SA NCERT Misc. , Q.12 , Page 52

Question

$\frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3} = \frac{9}{4}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}$

Step-by-step Solution

$\frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3}$

$= \frac{9}{4}\left[ {\frac{\pi }{2} - {{\sin }^{ - 1}}\frac{1}{3}} \right] = \frac{9}{4}{\cos ^{ - 1}}\frac{1}{3} = \frac{9}{4}{\sin ^{ - 1}}\left( {\frac{{2\sqrt 2 }}{3}} \right)$

Hence, L.H.S. = R.H.S.

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