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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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Correct option is B

Put x=sin⁡θ then sin−1⁡x2+1−x22 =sin−1⁡12sin⁡θ+12cos⁡θ=sin−1⁡sin⁡θ+π4=θ+π4∴given expression =tan⁡θ+π4−θ=tan⁡π4=1

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

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