JEE Advanced 2025 · Paper 2, Mathematics question 3

Let $t=\tan\theta$. Note that $\frac{6t}{9+t^2}=\sin\left(2\tan^{-1}\frac t3\right)$.

Case $|t|\le3$. Here $\frac12\sin^{-1}(\cdots)=\tan^{-1}\frac t3$, so
$$\theta=\tan^{-1}2t-\tan^{-1}\frac t3.$$
Taking tan of both sides: $t=\frac{5t/3}{1+2t^2/3}$, so $t=0$ or $t^2=1$.

• $t=0$ gives $\theta=0$.
• $t=\pm1$ gives right side $\pm\frac\pi4$, so $\theta=\pm\frac\pi4$.

These are 3 solutions, and only $n=0$ works in each family, because the right side is fixed.

Case $|t|>3$. The right side becomes $\tan^{-1}2t+\tan^{-1}\frac t3-\frac\pi2$ (for $t>3$; symmetrically for $t<-3$). Its tangent is $\frac{2t^2-3}{7t}$. Setting this equal to $t$ gives $5t^2=-3$, which has no solution.

Total: 3.