Given 0≤x≤12 then the value of tan sin-1 x2+1-x22-sin-1 x is

tan⁡sin−1⁡x2+1−x22−sin−1⁡x=tan⁡sin−1⁡x+1−x22−sin−1⁡x Put sin−1⁡x=θ⇒x=sin⁡θ=tan⁡sin−1⁡sin⁡θ+1−sin2⁡θ2−θ=tan⁡sin−1⁡12sin⁡θ+12cos⁡θ−θ=tan⁡sin−1⁡sin⁡θ+π4−θ=tan⁡θ+π4−θ=tan⁡π4=1

Given 0≤x≤12 then the value of tan sin-1 x2+1-x22-sin-1 x is

tan⁡sin−1⁡x2+1−x22−sin−1⁡x=tan⁡sin−1⁡x+1−x22−sin−1⁡x Put sin−1⁡x=θ⇒x=sin⁡θ=tan⁡sin−1⁡sin⁡θ+1−sin2⁡θ2−θ=tan⁡sin−1⁡12sin⁡θ+12cos⁡θ−θ=tan⁡sin−1⁡sin⁡θ+π4−θ=tan⁡θ+π4−θ=tan⁡π4=1

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Given $0 \leq x \leq \frac{1}{2}$ then the value of $\tan [\sin^{- 1} \{\frac{x}{\sqrt{2}} + \frac{\sqrt{1 - x^{2}}}{\sqrt{2}}\} - \sin^{- 1} x]$ is

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-1

$\frac{1}{\sqrt{3}}$

$\sqrt{3}$

answer is B.

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Detailed Solution

$\begin{aligned} \tan [\sin^{- 1} \{\frac{x}{\sqrt{2}} + \frac{\sqrt{1 - x^{2}}}{\sqrt{2}}\} - \sin^{- 1} x] \\ = \tan [\sin^{- 1} \{\frac{x + \sqrt{1 - x^{2}}}{\sqrt{2}}\} - \sin^{- 1} x] \end{aligned}$
$Put \sin^{- 1} x = θ \Rightarrow x = \sin θ$

$\begin{array}{l} = \tan [[\sin^{- 1} \{\frac{\sin θ + \sqrt{1 - \sin^{2} θ}}{\sqrt{2}}\} - θ] \\ = \tan [\sin^{- 1} \{\frac{1}{\sqrt{2}} \sin θ + \frac{1}{\sqrt{2}} \cos θ\} - θ] \\ = \tan [\sin^{- 1} \{\sin (θ + \frac{π}{4})\} - θ] \\ = \tan [θ + \frac{π}{4} - θ] \\ = \tan \frac{π}{4} = 1 \end{array}$