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