^ - 1 3 4 + ^ - 1 5 13 = - Vidyadip

MCQ

${\cot ^{ - 1}}\frac{3}{4} + {\sin ^{ - 1}}\frac{5}{{13}} = $

✓
${\sin ^{ - 1}}\frac{{63}}{{65}}$
B
${\sin ^{ - 1}}\frac{{12}}{{13}}$
C
${\sin ^{ - 1}}\frac{{65}}{{68}}$
D
${\sin ^{ - 1}}\frac{5}{{12}}$

✓

Answer

Correct option: A.

${\sin ^{ - 1}}\frac{{63}}{{65}}$

a
(a) Let ${\cot ^{ - 1}}\frac{3}{4} = \theta \,\, \Rightarrow \,\,\cot \theta = \frac{3}{4}$

and $\sin \theta = \frac{1}{{\sqrt {1 + {{\cot }^2}\theta } }} = \frac{1}{{\sqrt {1 + (9/16)} }} = \frac{4}{5}$

Hence ${\cot ^{ - 1}}\frac{3}{4} + {\sin ^{ - 1}}\frac{5}{{13}} = {\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{5}{{13}}$

$ = {\sin ^{ - 1}}\left[ {\frac{4}{5}.\sqrt {1 - \frac{{25}}{{169}}} + \frac{5}{{13}}.\,\sqrt {1 - \frac{{16}}{{25}}} } \right]$

$ = {\sin ^{ - 1}}\left[ {\frac{4}{5}.\frac{{12}}{{13}} + \frac{5}{{13}}.\frac{3}{5}} \right]$

$ = {\sin ^{ - 1}}\left[ {\frac{{48 + 15}}{{65}}} \right] = {\sin ^{ - 1}}\frac{{63}}{{65}}$.

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