KCET · Maths · Application of Derivatives
Given \( 0 \leq x \leq \frac{1}{2} \) then the value of
\[
\tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right] \text { is }
\]
- A \( 0 \sqrt{3} \)
- B \( \frac{1}{\sqrt{3}} \)
- C 1
- D \( -1 \)
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
Given that
\[
\begin{array}{l}
\tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right] \\
=\tan \left[\sin ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right] \\
\text { Let } \sin ^{-1} x=\theta \Rightarrow x=\sin \theta \\
\text { Now, } \tan \left[\sin ^{-1}\left\{\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right\}-\theta\right] \\
=\tan \left[\sin ^{-1}\left\{\sin \left(\theta+\frac{\Pi}{4}\right)-\theta\right\}\right] \\
=\tan \left[\theta+\frac{\Pi}{4}-\theta\right]=\tan \frac{\Pi}{4}=1
\end{array}
\]
\[
\begin{array}{l}
\tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right] \\
=\tan \left[\sin ^{-1}\left\{\frac{x+\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right] \\
\text { Let } \sin ^{-1} x=\theta \Rightarrow x=\sin \theta \\
\text { Now, } \tan \left[\sin ^{-1}\left\{\frac{\sin \theta+\cos \theta}{\sqrt{2}}\right\}-\theta\right] \\
=\tan \left[\sin ^{-1}\left\{\sin \left(\theta+\frac{\Pi}{4}\right)-\theta\right\}\right] \\
=\tan \left[\theta+\frac{\Pi}{4}-\theta\right]=\tan \frac{\Pi}{4}=1
\end{array}
\]
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