Question

The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)

• A. 1
• B. 2
• C. 3
• D. 5

Hint:

When solving equations involving inverse trigonometric functions, substitution and domain analysis are key. Also, symmetry and graphical analysis can quickly give the count of solutions.

Answer 1

The Correct Option is C

Step 1: Let \( x = \tan \theta \). Then \( \theta = \tan^{-1} x \) Substitute in the given equation: \[ \tan^{-1} x = \tan^{-1}(2x) - \frac{1}{2} \sin^{-1} \left( \frac{6x}{9 + x^2} \right) \] 
Step 2: Apply \( \tan^{-1} x = \theta \Rightarrow x = \tan \theta \) and the domain restriction \( \theta \in (-\frac{\pi}{2}, \frac{\pi}{2}) \Rightarrow x \in \mathbb{R} \) We rewrite the equation: \[ \tan^{-1} x = \tan^{-1}(2x) - \frac{1}{2} \sin^{-1} \left( \frac{6x}{9 + x^2} \right) \] 
Step 3: Define LHS = RHS and analyze the function: Let \[ f(x) = \tan^{-1} x - \tan^{-1}(2x) + \frac{1}{2} \sin^{-1} \left( \frac{6x}{9 + x^2} \right) \] We want to solve \( f(x) = 0 \) 
Step 4: Observe that \[ \tan^{-1} x - \tan^{-1}(2x) = \tan^{-1} \left( \frac{x - 2x}{1 + 2x^2} \right) = \tan^{-1} \left( \frac{-x}{1 + 2x^2} \right) \] So, \[ f(x) = \tan^{-1} \left( \frac{-x}{1 + 2x^2} \right) + \frac{1}{2} \sin^{-1} \left( \frac{6x}{9 + x^2} \right) \] Step 5: Let us analyze \( f(x) = 0 \Rightarrow \) \[ \tan^{-1} \left( \frac{-x}{1 + 2x^2} \right) = -\frac{1}{2} \sin^{-1} \left( \frac{6x}{9 + x^2} \right) \] So define a function: \[ g(x) = \tan^{-1} \left( \frac{-x}{1 + 2x^2} \right) + \frac{1}{2} \sin^{-1} \left( \frac{6x}{9 + x^2} \right) \] We now find the number of real solutions of \( g(x) = 0 \) 
Step 6: Graphical/Monotonicity Analysis (or plotting): Using symmetry and bounds: - \( \left| \frac{-x}{1 + 2x^2} \right| \leq \frac{1}{2\sqrt{2}} \Rightarrow \tan^{-1} \text{ bounded} \) - \( \left| \frac{6x}{9 + x^2} \right| \leq 1 \Rightarrow \sin^{-1} \text{ defined} \)
Graphing \( g(x) \) shows that it crosses the x-axis three times. So, the number of real solutions is 3.