JEE Advanced 2025 — Paper 2 — Question 19

Let $t = \tan \theta$. Then $\theta = \tan^{-1} t$ (since $\theta \in (-\pi/2, \pi/2)$). The equation becomes $\tan^{-1} t = \tan^{-1}(2t) - \frac{1}{2} \sin^{-1}\left(\frac{6t}{9+t^2}\right)$. Set $\alpha = \frac{1}{2} \sin^{-1}\left(\frac{6t}{9+t^2}\right)$, so $\sin 2\alpha = \frac{6t}{9+t^2}$. Using the identity $\sin 2\alpha = \frac{2 \tan \alpha}{1+\tan^2 \alpha}$, we get $\frac{2 \tan \alpha}{1+\tan^2 \alpha} = \frac{6t}{9+t^2}$. From the equation, $\tan^{-1} t + \alpha = \tan^{-1}(2t)$. Taking tangent both sides: $\tan(\tan^{-1} t + \alpha) = 2t \Rightarrow \frac{t + \tan \alpha}{1 - t \tan \alpha} = 2t$. Simplify: $t + \tan \alpha = 2t - 2t^2 \tan \alpha \Rightarrow 3t \tan \alpha + \tan \alpha = t \Rightarrow \tan \alpha (3t + 1) = t$. (Note: This step differs from the given solution; we need to re-derive carefully.) Actually, better approach: From $\frac{t + \tan \alpha}{1 - t \tan \alpha} = 2t$, cross-multiply: $t + \tan \alpha = 2t - 2t^2 \tan \alpha$. Rearrange: $\tan \alpha + 2t^2 \tan \alpha = t \Rightarrow \tan \alpha (1 + 2t^2) = t$. Thus $\tan \alpha = \frac{t}{1 + 2t^2}$. Now substitute into $\sin 2\alpha = \frac{6t}{9+t^2}$. Write $\sin 2\alpha = \frac{2 \tan \alpha}{1+\tan^2 \alpha}$. Substitute $\tan \alpha = \frac{t}{1+2t^2}$: $\frac{2 \cdot \frac{t}{1+2t^2}}{1 + \left(\frac{t}{1+2t^2}\right)^2} = \frac{6t}{9+t^2}$. Simplify numerator and denominator: Numerator: $\frac{2t}{1+2t^2}$. Denominator: $1 + \frac{t^2}{(1+2t^2)^2} = \frac{(1+2t^2)^2 + t^2}{(1+2t^2)^2} = \frac{1 + 4t^2 + 4t^4 + t^2}{(1+2t^2)^2} = \frac{1 + 5t^2 + 4t^4}{(1+2t^2)^2}$. So LHS = $\frac{2t}{1+2t^2} \cdot \frac{(1+2t^2)^2}{1+5t^2+4t^4} = \frac{2t(1+2t^2)}{1+5t^2+4t^4}$. Set equal to $\frac{6t}{9+t^2}$. If $t=0$, both sides are 0, giving $\theta = 0$. Otherwise, divide by $t$: $\frac{2(1+2t^2)}{1+5t^2+4t^4} = \frac{6}{9+t^2}$. Cross-multiply: $2(1+2t^2)(9+t^2) = 6(1+5t^2+4t^4)$. Expand LHS: $2[(1)(9) + 1(t^2) + 2t^2(9) + 2t^2(t^2)] = 2[9 + t^2 + 18t^2 + 2t^4] = 2[9 + 19t^2 + 2t^4] = 18 + 38t^2 + 4t^4$. RHS: $6 + 30t^2 + 24t^4$. Equate: $18 + 38t^2 + 4t^4 = 6 + 30t^2 + 24t^4$. Bring all to one side: $12 + 8t^2 - 20t^4 = 0$. Divide by 4: $3 + 2t^2 - 5t^4 = 0$. Let $u = t^2$: $5u^2 - 2u - 3 = 0 \Rightarrow (5u+3)(u-1)=0$. So $u = 1$ or $u = -3/5$ (discard). Thus $t^2 = 1 \Rightarrow t = \pm 1$. Corresponding $\theta = \pi/4$ and $\theta = -\pi/4$. Together with $t=0$, we have three solutions: $\theta = 0,\ \pi/4,\ -\pi/4$. The total number of real solutions is 3.