JEE Advanced 2025 Paper 2Question 1
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 $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, respectively.)
Inverse Trigonometric Functions
Solution
$|\text{RHS}|<\frac{3\pi}4$; with $t=\tan\theta$ and $|t|\le3$, $\frac12\sin^{-1}\frac{6t}{9+t^2}=\tan^{-1}\frac t3$, so for $|\theta|<\frac\pi2$: $\tan\theta=\frac{2t-\frac t3}{1+\frac{2t^2}3}=\frac{5t}{3+2t^2}=t$, giving $t=0,\pm1$. The branches $\frac\pi2<|\theta|<\frac{3\pi}4$ give no solution, so $\theta=0,\pm\frac\pi4$: $3$ solutions.