MHT CET · Maths · Inverse Trigonometric Functions
The value of \(\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)\) at \(x=\frac{1}{5}\), where \(0 \leq \cos ^{-1} x \leq \pi\) and \(-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}\), is
- A \(-\frac{\sqrt{6}}{5}\)
- B \(\frac{\sqrt{6}}{5}\)
- C \(-\frac{2 \sqrt{6}}{5}\)
- D \(\frac{2 \sqrt{6}}{5}\)
Answer & Solution
Correct Answer
(C) \(-\frac{2 \sqrt{6}}{5}\)
Step-by-step Solution
Detailed explanation
\(\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)=\cos\) \(\left(\cos ^{-1} x+\cos ^{-1}x+\sin ^{-1} x\right) \)
\( =\cos \left(\cos ^{-1} x+\frac{\pi}{2}\right) \quad\left[\because \cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}\right] \)
\( =-\sin \left(\cos ^{-1} x\right) \)
\( =-\sin \left(\sin ^{-1} \sqrt{1-x^2}\right) \)
\( =-\sqrt{1-x^2} \)
\( =-\sqrt{1-\left(\frac{1}{5}\right)^2} \quad\left[\because x=\frac{1}{5}\right] \)
\( =-\frac{2 \sqrt{6}}{5}\)
\( =\cos \left(\cos ^{-1} x+\frac{\pi}{2}\right) \quad\left[\because \cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}\right] \)
\( =-\sin \left(\cos ^{-1} x\right) \)
\( =-\sin \left(\sin ^{-1} \sqrt{1-x^2}\right) \)
\( =-\sqrt{1-x^2} \)
\( =-\sqrt{1-\left(\frac{1}{5}\right)^2} \quad\left[\because x=\frac{1}{5}\right] \)
\( =-\frac{2 \sqrt{6}}{5}\)
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