Question

The simplified expression of \(\sin(\tan^{-1}x)\), for any real number \(x\) is given by

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

Hint:

If \(\theta=\tan^{-1}x\), build triangle with opposite \(x\), adjacent \(1\), hypotenuse \(\sqrt{1+x^2}\).

Answer 1

The Correct Option is B

Step 1: Let \(\theta = \tan^{-1}x\).
Then:
\[ \tan\theta = x = \frac{\text{opposite}}{\text{adjacent}} \]
Step 2: Draw right triangle.
Take opposite side = \(x\), adjacent = \(1\).
So hypotenuse =
\[ \sqrt{x^2+1} \]
Step 3: Write \(\sin\theta\).
\[ \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{1+x^2}} \]
Final Answer:
\[ \boxed{\frac{x}{\sqrt{1+x^2}}} \]