MHT CET · Maths · Inverse Trigonometric Functions
If \(\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}\), then the value of \(x\) is
- A 4
- B 1
- C 5
- D 3
Answer & Solution
Correct Answer
(D) 3
Step-by-step Solution
Detailed explanation
\(\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2} \)
\( \Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)=\frac{\pi}{2} \)
\( \Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right) \)
\( \Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\cos ^{-1}\left(\frac{4}{5}\right) \quad \cdots[\because \sin ^{-1} x+\cos ^{-1}\) \(x=\frac{\pi}{2}] \)
\( \Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\sin ^{-1}\left(\frac{3}{5}\right) \cdots[\cos ^{-1} x=\sin ^{-1}\) \(\sqrt{1-x^2}] \)
\( \Rightarrow x=3\)
\( \Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)=\frac{\pi}{2} \)
\( \Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right) \)
\( \Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\cos ^{-1}\left(\frac{4}{5}\right) \quad \cdots[\because \sin ^{-1} x+\cos ^{-1}\) \(x=\frac{\pi}{2}] \)
\( \Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\sin ^{-1}\left(\frac{3}{5}\right) \cdots[\cos ^{-1} x=\sin ^{-1}\) \(\sqrt{1-x^2}] \)
\( \Rightarrow x=3\)
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