Trigonometry – Inverse Trigonometric Functions – JEE Main 28 Jan 2026 Shift 1

Solution:

Let $\cos^{-1}\frac{2}{3} = 2\alpha$. Then $\sin^{-1}\frac{2}{3} = \frac{\pi}{2} – 2\alpha$.
$$k = \tan\left(\frac{\pi}{4} + \alpha\right) + \tan\left(\frac{1}{2}\left(\frac{\pi}{2} – 2\alpha\right)\right)$$
$$k = \tan\left(\frac{\pi}{4} + \alpha\right) + \tan\left(\frac{\pi}{4} – \alpha\right)$$
$$k = \frac{1+\tan\alpha}{1-\tan\alpha} + \frac{1-\tan\alpha}{1+\tan\alpha} = \frac{(1+\tan\alpha)^2 + (1-\tan\alpha)^2}{1-\tan^2\alpha}$$
$$k = \frac{2(1+\tan^2\alpha)}{1-\tan^2\alpha} = \frac{2}{\cos 2\alpha}$$
Since $\cos 2\alpha = \frac{2}{3}$, we have $k = \frac{2}{2/3} = 3$.

Now solve the equation: $\sin^{-1}(3x-1) = \sin^{-1}x – \cos^{-1}x$.
Using $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$, we substitute $\cos^{-1}x = \frac{\pi}{2} – \sin^{-1}x$:
$$\sin^{-1}(3x-1) = \sin^{-1}x – \left(\frac{\pi}{2} – \sin^{-1}x\right)$$
$$\sin^{-1}(3x-1) = 2\sin^{-1}x – \frac{\pi}{2}$$

Let $\sin^{-1}x = \theta$. Then $x = \sin\theta$ and $\theta \in [-\pi/2, \pi/2]$.
$$3x – 1 = \sin\left(2\theta – \frac{\pi}{2}\right)$$
$$3x – 1 = -\cos(2\theta) = -(1 – 2\sin^2\theta)$$
$$3x – 1 = 2x^2 – 1$$
$$2x^2 – 3x = 0 \Rightarrow x(2x – 3) = 0$$
Possible solutions: $x = 0$ or $x = 1.5$.

Since $x$ is in the domain of $\sin^{-1}x$, $x \in [-1, 1]$. Thus, $x = 1.5$ is rejected.
Check $x = 0$:
LHS: $\sin^{-1}(-1) = -\frac{\pi}{2}$.
RHS: $\sin^{-1}(0) – \cos^{-1}(0) = 0 – \frac{\pi}{2} = -\frac{\pi}{2}$.
LHS = RHS.

So, $x = 0$ is the only solution.
Number of solutions is 1.

Ans. 1