Question

The value of
\[ \sin \left[ \tan^{-1} \left( \frac{1 - x^2}{2x} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \right] \]

• A. 0
• B. 1
• C. -1
• D. \( \frac{1}{2} \)

Hint:

When dealing with the sum of inverse trigonometric functions, you can use the identity for the sine of a sum to simplify the expression.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to simplify an expression containing inverse trigonometric functions inside a sine function.
Step 2: Key Formula or Approach:
Use trigonometric substitution \( x = \tan \theta \).
Step 3: Detailed Explanation:
1. Let \( x = \tan \theta \). Then:
\( \frac{1-x^2}{2x} = \frac{1-\tan^2 \theta}{2\tan \theta} = \frac{1}{\tan 2\theta} = \cot 2\theta \).
\( \tan^{-1}(\cot 2\theta) = \tan^{-1}(\tan(\pi/2 - 2\theta)) = \pi/2 - 2\theta \).
2. Now simplify the second term:
\( \frac{1-x^2}{1+x^2} = \cos 2\theta \).
\( \cos^{-1}(\cos 2\theta) = 2\theta \).
3. Combine the terms:
\( \text{Sum} = (\pi/2 - 2\theta) + (2\theta) = \pi/2 \).
4. Evaluate the outer function:
\( \sin(\pi/2) = 1 \).
Step 4: Final Answer:
The value of the expression is 1.