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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Advanced 2025 · Paper 2 · Mathematics · Inverse Trigonometric Functions

The total number of real solutions of the equation θ= ^ -1 (2...

The total number of real solutions of the equation θ=tan⁡−1(2tan⁡θ)−12sin⁡−1(6tan⁡θ9+tan⁡2θ)\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⁡−1xand\sin ^{-1} x and\tan ^{-1} xassumevaluesin[−π2,π2]and(−π2,π2)assume values in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] and \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), respectively.)

Options

  1. A1
  2. B2
  3. C3
  4. D5

Correct answer

C. 3

Step-by-step solution

Let α=12sin⁡−1(6tan⁡θ9+tan⁡2θ)\alpha=\frac{1}{2} \sin ^{-1}\left(\frac{6 \tan \theta}{9+\tan ^2 \theta}\right)θ=tan⁡−1(2tan⁡θ)−αθ+α=tan⁡−1(2tan⁡θ)tan⁡(θ+α)=2tan⁡θtan⁡θ+tan⁡α1−tan⁡αtan⁡θ=2tan⁡θ....(1)sin⁡2α=6tan⁡θ9+tan⁡2θ=2tan⁡α1+tan⁡2α3tan⁡θ+3tan⁡θtan⁡2α=9tan⁡α+tan⁡αtan⁡2θ3(tan⁡θ−3tan⁡α)=tan⁡αtan⁡θ(tan⁡θ−3tan⁡α)tan⁡θ=3tan⁡α or  tan⁡θ=3tan⁡α\begin{aligned} & \theta=\tan ^{-1}(2 \tan \theta)-\alpha \\ & \theta+\alpha=\tan ^{-1}(2 \tan \theta) \\ & \tan (\theta+\alpha)=2 \tan \theta \\ & \frac{\tan \theta+\tan \alpha}{1-\tan \alpha \tan \theta}=2 \tan \theta ....(1)\\ & \sin 2 \alpha=\frac{6 \tan \theta}{9+\tan ^2 \theta}=\frac{2 \tan \alpha}{1+\tan ^2 \alpha} \\ & 3 \tan \theta+3 \tan \theta \tan ^2 \alpha=9 \tan \alpha+\tan \alpha \tan ^2 \theta \\ & 3(\tan \theta-3 \tan \alpha)=\tan \alpha \tan \theta(\tan \theta-3 \tan \alpha) \\ & \tan \theta=\frac{3}{\tan \alpha} \text{\text{ or } } \tan \theta=3 \tan \alpha \end{aligned}$\begin{aligned} & \text{ Case-I : } \tan \theta=3 \tan \alpha \\ & \frac{\tan \theta+\frac{\tan \theta}{3}}{1-\frac{\tan ^2 \theta}{3}}=2 \tan \theta\end{aligned}\tan \theta=0, \quad \frac{2}{3}=1-\frac{\tan ^2 \theta}{3} \quad \Rightarrow \tan \theta=1,-1\tan \theta=0,-1,1 \\  Case-II : tan⁡θ=3tan⁡αtan⁡θ+3tan⁡θ−2=2tan⁡θ\begin{aligned}& \text{ Case-II : } \tan \theta=\frac{3}{\tan \alpha}\\ & \frac{\tan \theta+\frac{3}{\tan \theta}}{-2}=2 \tan \theta\end{aligned}$
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