MCQ
sin (tan-1 x), |x| < 1 का मान बराबर है:
  • A
    $\frac{1}{\sqrt{1-x^{2}}}$
  • B
    $\frac{x}{\sqrt{1+x^{2}}}$
  • C
    $\frac{1}{\sqrt{1+x^{2}}}$
  • D
    $\frac{x}{\sqrt{1-x^{2}}}$

Answer

(sin tan-1 x) = sin [sin-1 $\frac{x}{\sqrt{1+x^{2}}}$] ($\because$ tan-1 x = sin-1 $\frac{x}{\sqrt{1+x^{2}}}$)
= $\frac{x}{\sqrt{1+x^{2}}}$

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