Q67 / 180 · KCET 2014 · Mathematics
Given \( 0 \leq x \leq \frac{1}{2} \) then the value of
\[
\tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right] \text { is }
\]
\[ \tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right] \text { is } \]
- A\( 0 \sqrt{3} \)
- B\( \frac{1}{\sqrt{3}} \)
- Correct answerC1
- D\( -1 \)
Solution
Given that
\[ \begin{array}{l} \tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right] \\ =\tan \left[\sin ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right] \\ \text { Let } \sin ^{-1} x=\theta \Rightarrow x=\sin \theta \\ \text { Now, } \tan \left[\sin ^{-1}\left\{\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right\}-\theta\right] \\ =\tan \left[\sin ^{-1}\left\{\sin \left(\theta+\frac{\Pi}{4}\right)-\theta\right\}\right] \\ =\tan \left[\theta+\frac{\Pi}{4}-\theta\right]=\tan \frac{\Pi}{4}=1 \end{array} \]
\[ \begin{array}{l} \tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right] \\ =\tan \left[\sin ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right] \\ \text { Let } \sin ^{-1} x=\theta \Rightarrow x=\sin \theta \\ \text { Now, } \tan \left[\sin ^{-1}\left\{\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right\}-\theta\right] \\ =\tan \left[\sin ^{-1}\left\{\sin \left(\theta+\frac{\Pi}{4}\right)-\theta\right\}\right] \\ =\tan \left[\theta+\frac{\Pi}{4}-\theta\right]=\tan \frac{\Pi}{4}=1 \end{array} \]
Hence, the correct answer is option: 'C'.