Question

Prove that \( \sin^{-1} x = \tan^{-1} [x / \sqrt{1-x^2}] \).

Hint:

Drawing a simple right-angled triangle is the fastest and most intuitive way to handle conversions between inverse trigonometric functions. This geometric approach helps avoid algebraic mistakes.

Answer 1

Step 1: Understanding the Concept:
We can prove this identity by converting one inverse trigonometric function to another. A common method is to use a right-angled triangle to represent the relationship given by the first function and then derive the second function from the triangle's dimensions.
Step 2: Key Formula or Approach:
1. Let \( \theta = \sin^{-1} x \). This implies \( \sin \theta = x \).
2. Construct a right-angled triangle where \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{x}{1} \).
3. Use the Pythagorean theorem to find the length of the adjacent side.
4. From the triangle, find \( \tan \theta \) and then express \( \theta \) in terms of \( \tan^{-1} \).
Step 3: Detailed Explanation:
Let \( \theta = \sin^{-1} x \). Then, by definition, \( \sin \theta = x \). We can write this as \( \sin \theta = \frac{x}{1} \).
Consider a right-angled triangle with angle \( \theta \). We can set:
Opposite side = \( x \)
Hypotenuse = \( 1 \)
Using the Pythagorean theorem (\( \text{Opposite}^2 + \text{Adjacent}^2 = \text{Hypotenuse}^2 \)): \[ x^2 + (\text{Adjacent})^2 = 1^2 \] \[ (\text{Adjacent})^2 = 1 - x^2 \] \[ \text{Adjacent} = \sqrt{1 - x^2} \] Now, we can find the value of \( \tan \theta \) from this triangle: \[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{x}{\sqrt{1 - x^2}} \] Taking the inverse tangent of both sides, we get: \[ \theta = \tan^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right) \] Since we started with \( \theta = \sin^{-1} x \), we have proven that: \[ \sin^{-1} x = \tan^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right) \] Step 4: Final Answer:
The identity is proven by equating the two expressions for \( \theta \).