Question

\(\sin(\tan^{-1}x),\ |x|<1\) is equal to:

• A. \(\dfrac{x}{\sqrt{1-x^2}}\)
• B. \(\dfrac{1}{\sqrt{1-x^2}}\)
• C. \(\dfrac{1}{\sqrt{1+x^2}}\)
• D. \(\dfrac{x}{\sqrt{1+x^2}}\)

Hint:

Draw the reference right triangle for \(\theta=\tan^{-1}x\) and read off \(\sin\theta\).

Answer 1

The Correct Option is D

Step 1: Using the identity directly:
With \(\theta=\tan^{-1}x\), \(\sec^2\theta=1+\tan^2\theta=1+x^2\), so \(\cos\theta=\dfrac{1}{\sqrt{1+x^2}}\) (principal branch, \(\cos\theta>0\)).

Step 2: Getting sine from cosine:
\(\sin\theta=\tan\theta\cdot\cos\theta=x\cdot\dfrac{1}{\sqrt{1+x^2}}=\dfrac{x}{\sqrt{1+x^2}}\).

Final Answer:
Confirmed: \(\boxed{\dfrac{x}{\sqrt{1+x^2}}}\).