JEE Advanced 2025 Paper 2 Q03 Mathematics Trigonometry Inverse Trigonometric Functions Hard

JEE Advanced 2025 Paper 2 · Q03 · Inverse Trigonometric Functions

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 $[-\pi/2, \pi/2]$ and $(-\pi/2, \pi/2)$, respectively.)

  1. A. 1
  2. B. 2
  3. C. 3
  4. D. 5
Reveal answer + step-by-step solution

Correct answer:C

Solution

Let $u = \tan(\theta)$. Note $\frac{6u}{9 + u^2} = \frac{2(u/3)}{1 + (u/3)^2}$. Let $\phi = \arctan(u/3)$ so that this expression equals $\sin(2\phi)$. For $|u| \le 3$, $2\phi \in [-\pi/2, \pi/2]$ and $\frac{1}{2} \sin^{-1}(\ldots) = \phi = \arctan(u/3)$. The equation becomes $\theta = \arctan(2u) - \arctan(u/3)$. Using $\arctan A - \arctan B = \arctan\left(\frac{A-B}{1+AB}\right)$ (valid for $1+AB > 0$): $= \arctan\left(\frac{2u - u/3}{1 + 2u^2/3}\right) = \arctan\left(\frac{5u}{3 + 2u^2}\right)$. So $\tan(\theta) = \frac{5u}{3 + 2u^2}$. Substitute $u = \tan(\theta)$: $u(3 + 2u^2) = 5u$ $\to$ $u(2u^2 - 2) = 0$ $\to$ $u = 0$ or $u = \pm 1$, giving $\theta = 0, \pm \pi/4$. Three solutions in the principal range, and the branches $|u|>3$ yield no further solutions (the range condition pulls the candidates back). Answer: $3$, option (C).

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