Question

sin tan^-1x, |x| < 1 is equal to
(A) \(\frac{x}{\sqrt{1-x^{2}}}\)
(B) \(\frac{1}{\sqrt{1-x^{2}}}\)
(C) \(\frac{1}{\sqrt{1+x^{2}}}\)
(D) \(\frac{x}{\sqrt{1+x^{2}}}\)

Answer 1

Let tan^-1x = α ∴ tan α = x.
So, sin α = \(\frac{x}{\sqrt{1+x^{2}}}\)
or α = sin^-1\(\frac{x}{\sqrt{1+x^{2}}}\)
Now sin tan^-1x = sin α = sin(sin^-1\(\frac{x}{\sqrt{1+x^{2}}}\))
= \(\frac{x}{\sqrt{1+x^{2}}}\)