MHT CET · Maths · Inverse Trigonometric Functions
Given \(0 \leq x \leq \frac{1}{2}\), then the value of \(\tan \left(\sin ^{-1}\left(\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^2}}{\sqrt{2}}\right)-\sin ^{-1} x\right)\) is
- A \(1\)
- B \(\sqrt 3\)
- C \(-1\)
- D \(\frac{1}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
\( \tan \left[\sin ^{-1}\left(\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^2}}{\sqrt{2}}\right)-\sin ^{-1} x\right] \)
\( =\tan \left[\sin ^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)-\sin ^{-1} x\right] \)
\( =\tan \left[\sin ^{-1}\left(\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right)-\theta\right]\) \(\ldots[\text { Put } \sin ^{-1} x=\theta\)
\( \Rightarrow x=\sin \theta] \)
\( =\tan \left[\sin ^{-1}\left[\sin \left(\theta+\frac{\pi}{4}\right)\right]-\theta\right] \)
\( =\tan \left(\theta+\frac{\pi}{4}-\theta\right) \)
\( =\tan \frac{\pi}{4}=1\)
\( =\tan \left[\sin ^{-1}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)-\sin ^{-1} x\right] \)
\( =\tan \left[\sin ^{-1}\left(\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right)-\theta\right]\) \(\ldots[\text { Put } \sin ^{-1} x=\theta\)
\( \Rightarrow x=\sin \theta] \)
\( =\tan \left[\sin ^{-1}\left[\sin \left(\theta+\frac{\pi}{4}\right)\right]-\theta\right] \)
\( =\tan \left(\theta+\frac{\pi}{4}-\theta\right) \)
\( =\tan \frac{\pi}{4}=1\)
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