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

The Correct Option is D

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