Question

Solve: $sin(tan^{-1}x)=|x|<1$ is equal to

• A. $\frac {x}{\sqrt{1-x^2}}$
• B. $\frac {1}{\sqrt{1-x^2}}$
• C. $\frac {x}{\sqrt{1+x^2}}$
• D. $\frac {x}{\sqrt{1+x^2}}$

Answer 1

Answer: (D)

$\frac {x}{\sqrt{1+x^2}}$

Answer 2

Let tan−1x = y.
Then, tan y = x$\implies$sin y = $\frac {x}{\sqrt{1+x^2}}$
y = sin-1\frac {x}{\sqrt{1+x^2}}$$\impliestan-1x = sin-1$\frac {x}{\sqrt{1+x^2}}$
Therefore, sin(tan-1x) = sin(sin-1$\frac {x}{\sqrt{1+x^2}}$) =$\frac {x}{\sqrt{1+x^2}}$

Hence, the correct answer is (D): $\frac {x}{\sqrt{1+x^2}}$