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: Setting up a right triangle:
Let \(\theta=\tan^{-1}x\), so \(\tan\theta=x=\dfrac{\text{opposite}}{\text{adjacent}}\). Take opposite \(=x\), adjacent \(=1\).

Step 2: Finding the hypotenuse:
Hypotenuse \(=\sqrt{x^2+1}\) by Pythagoras.

Step 3: Reading off sine:
\(\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{x}{\sqrt{1+x^2}}\).

Final Answer:
\(\sin(\tan^{-1}x)=\boxed{\dfrac{x}{\sqrt{1+x^2}}}\).