Question

for $|x| < 1$, sin(tan-1x) equal to

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

Hint:

Drawing a right-angled triangle is a very effective and quick method for simplifying expressions involving compositions of trigonometric and inverse trigonometric functions. Always label the sides based on the given inverse function and then use Pythagoras' theorem to find the third side.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This problem involves finding the value of a trigonometric function of an inverse trigonometric function. A common method is to use a right-angled triangle to represent the inverse trigonometric function and then find the required trigonometric ratio.
Step 2: Key Formula or Approach:
Let \(\theta = \tan^{-1}x\). This implies \(\tan\theta = x\). We need to find \(\sin\theta\).
We can visualize this relationship using a right-angled triangle.
Step 3: Detailed Explanation:
Let \(\theta = \tan^{-1}x\). Then \(\tan\theta = x\).
We can write \(\tan\theta = \frac{x}{1}\).
In a right-angled triangle, \(\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}}\).
So, we can let the side opposite to angle \(\theta\) be \(x\) and the adjacent side be 1.
Using the Pythagorean theorem, we can find the hypotenuse:
\[ \text{Hypotenuse}^2 = \text{Opposite}^2 + \text{Adjacent}^2 \] \[ \text{Hypotenuse}^2 = x^2 + 1^2 = 1 + x^2 \] \[ \text{Hypotenuse} = \sqrt{1 + x^2} \] Now, we need to find \(\sin(\tan^{-1}x)\), which is \(\sin\theta\).
The formula for \(\sin\theta\) is \(\frac{\text{Opposite}}{\text{Hypotenuse}}\).
\[ \sin\theta = \frac{x}{\sqrt{1+x^2}} \] Step 4: Final Answer:
Therefore, \(\sin(\tan^{-1}x) = \frac{x}{\sqrt{1+x^2}}\).