MCQ
If ${\sin ^{ - 1}}\frac{x}{5} + {\rm{cose}}{{\rm{c}}^{ - 1}}\left( {\frac{5}{4}} \right) = \frac{\pi }{2},$ then $x = $
- A$4$
- B$5$
- C$1$
- ✓$3$
✓
Answer
Correct option: D.
$3$
d
Given $\sin ^{-1}\left(\frac{x}{5}\right)+\csc ^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$
Given $\sin ^{-1}\left(\frac{x}{5}\right)+\csc ^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right)$
$\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\cos ^{-1}\left(\frac{4}{5}\right)$
$\Rightarrow \frac{x}{5}=\sin \left(\cos ^{-1} \frac{4}{5}\right)$
$\Rightarrow \frac{x}{5}=\sin \left(\sin ^{-1} \frac{3}{5}\right)$
$\Rightarrow \frac{x}{5}=\frac{3}{5}$
$\Rightarrow x=3$
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