Question

The number of solutions of \( \tan^{-1} 1 + \frac{1}{2} \cos^{-1} x^2 - \tan^{-1}\left(\frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}\right) = 0 \) is

• A. $ 3 $
• B. $ 0 $
• C. $ 1 $
• D. infinitely many ( $\infty$ )

Hint:

When solving equations involving inverse trigonometric functions, consider their domains, ranges, and periodicity. Analyze how the terms interact to determine the number of solutions.

Answer 1

The Correct Option is D

Step 1: Analyze the given equation.
We are given: $$ \tan^{-1} 1 + \frac{1}{2} \cos^{-1} x^2 - \tan^{-1}\left(\frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}\right) = 0. $$ First, simplify each term: 1. Simplify \( \tan^{-1} 1 \): Since \( \tan^{-1} 1 = \frac{\pi}{4} \), we have: $$ \tan^{-1} 1 = \frac{\pi}{4}. $$ 2. Simplify \( \frac{1}{2} \cos^{-1} x^2 \): 
Let \( \theta = \cos^{-1} x^2 \). Then:
$$ \cos \theta = x^2 \quad \Rightarrow \quad \theta = \cos^{-1} x^2. $$ Thus: $$ \frac{1}{2} \cos^{-1} x^2 = \frac{1}{2} \theta. $$ 3. Simplify \( \tan^{-1}\left(\frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}\right) \): Let: $$ t = \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}. $$ Rationalize the denominator: $$ t = \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \cdot \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}} = \frac{(\sqrt{1+x^2} + \sqrt{1-x^2})^2}{(1+x^2) - (1-x^2)}. $$ Simplify the numerator and denominator: $$ (\sqrt{1+x^2} + \sqrt{1-x^2})^2 = (1+x^2) + 2\sqrt{(1+x^2)(1-x^2)} + (1-x^2) = 2 + 2\sqrt{1-x^4}, $$ $$ (1+x^2) - (1-x^2) = 2x^2. $$ Thus: $$ t = \frac{2 + 2\sqrt{1-x^4}}{2x^2} = \frac{1 + \sqrt{1-x^4}}{x^2}. $$ Therefore: $$ \tan^{-1}\left(\frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}\right) = \tan^{-1}\left(\frac{1 + \sqrt{1-x^4}}{x^2}\right). $$ Step 2: Substitute back into the equation.
The equation becomes: $$ \frac{\pi}{4} + \frac{1}{2} \cos^{-1} x^2 - \tan^{-1}\left(\frac{1 + \sqrt{1-x^4}}{x^2}\right) = 0. $$ Rearrange: $$ \tan^{-1}\left(\frac{1 + \sqrt{1-x^4}}{x^2}\right) = \frac{\pi}{4} + \frac{1}{2} \cos^{-1} x^2. $$ Step 3: Analyze the behavior of the equation.
The term \( \frac{1}{2} \cos^{-1} x^2 \) depends on \( x^2 \), which ranges from 0 to 1 as \( x \) varies over its domain. The term \( \tan^{-1}\left(\frac{1 + \sqrt{1-x^4}}{x^2}\right) \) also varies continuously with \( x \). Importantly, the equation involves trigonometric and inverse trigonometric functions, which can lead to infinitely many solutions due to periodicity and symmetry. 
Step 4: Conclusion.
Given the nature of the functions involved, the equation has infinitely many solutions. This is because the combination of \( \cos^{-1} x^2 \) and \( \tan^{-1}\left(\frac{1 + \sqrt{1-x^4}}{x^2}\right) \) allows for multiple values of \( x \) that satisfy the equation. 
Step 5: Final Answer.
$$ \boxed{\text{infinitely many} (\infty)}. $$