Question

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

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

Hint:

Draw a right triangle with opposite x, adjacent 1, hypotenuse \(\sqrt{1+x^2}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Let \( \theta=\tan^{-1}x \), so \( \tan\theta=x \) and \( \theta \) lies in \( \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right) \).
The task is to find \( \sin\theta \) in terms of x using a right triangle picture.

Step 2: Building the right triangle:
Since \( \tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{x}{1} \), take the opposite side as x and the adjacent side as 1.
By the Pythagorean theorem, the hypotenuse is \( \sqrt{x^2+1} \).

Step 3: Computing sine:
Sine is opposite over hypotenuse, so substitute the side lengths found above.
\[ \sin\theta=\dfrac{x}{\sqrt{1+x^2}} \]
This matches option A exactly.

Step 4: Why option B is wrong:
Option B, \( \dfrac{x}{\sqrt{1-x^2}} \), is actually the formula for \( \tan(\sin^{-1}x) \), not \( \sin(\tan^{-1}x) \); the sign under the root is wrong here.

Step 5: Why option C is wrong:
Option C, \( \dfrac{1}{\sqrt{1+x^2}} \), equals \( \cos(\tan^{-1}x) \), the adjacent over hypotenuse ratio, not the sine ratio.

Step 6: Why option D is wrong:
Option D, \( \dfrac{1}{\sqrt{1-x^2}} \), matches neither the sine nor the cosine of \( \tan^{-1}x \); it resembles a secant-type expression from a different substitution.

Final Answer:
Using the right triangle for \( \tan\theta=x \) gives sine directly. \[ \boxed{\sin(\tan^{-1}x)=\dfrac{x}{\sqrt{1+x^2}}} \]