Question:medium

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

Show 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.
Updated On: Mar 27, 2026
  • $\frac{1}{\sqrt{1+x^2}}$
  • $\frac{1}{\sqrt{1-x^2}}$
  • $\frac{x}{\sqrt{1-x^2}}$
  • $\frac{x}{\sqrt{1+x^2}}$
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Concept Review: This problem requires evaluating a trigonometric function applied to an inverse trigonometric function. The standard approach involves constructing a right-angled triangle to represent the inverse trigonometric function, from which the desired trigonometric ratio can be derived. Step 2: Core Methodology: Let \( \theta = \tan^{-1}x \). This definition implies that \( \tan\theta = x \). The objective is to determine the value of \( \sin\theta \). A right-angled triangle is an effective visual tool for this relationship. Step 3: In-depth Analysis: Given \( \theta = \tan^{-1}x \), it follows that \( \tan\theta = x \). This can be expressed as \( \tan\theta = \frac{x}{1} \). In the context of a right-angled triangle, \( \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} \). Consequently, we can assign the side opposite to angle \( \theta \) a length of \( x \) and the adjacent side a length of 1. The hypotenuse is calculated using the Pythagorean theorem: \[ \text{Hypotenuse}^2 = \text{Opposite}^2 + \text{Adjacent}^2 \] \[ \text{Hypotenuse}^2 = x^2 + 1^2 = 1 + x^2 \] \[ \text{Hypotenuse} = \sqrt{1 + x^2} \] The task is to find \( \sin(\tan^{-1}x) \), which is equivalent to \( \sin\theta \). The formula for \( \sin\theta \) is \( \frac{\text{Opposite}}{\text{Hypotenuse}} \). \[ \sin\theta = \frac{x}{\sqrt{1+x^2}} \] Step 4: Conclusion: Thus, the expression \( \sin(\tan^{-1}x) \) simplifies to \( \frac{x}{\sqrt{1+x^2}} \).
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