JEE Main 2026 — Inverse Trigonometric Functions Question with Solution

We are given $\alpha =3{\sin }^{-1}\left(\frac{6}{11}\right)$ and $\beta =3{\cos }^{-1}\left(\frac{4}{9}\right)$.

First, we find the range of $\alpha$.
Since $\frac{1}{2}<\frac{6}{11}<\frac{1}{\sqrt{2}}$, we have ${\sin }^{-1}\left(\frac{1}{2}\right)<{\sin }^{-1}\left(\frac{6}{11}\right)<{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)$.
$\Rightarrow \frac{\pi }{6}<{\sin }^{-1}\left(\frac{6}{11}\right)<\frac{\pi }{4}$
Multiplying by $3$, we get $\frac{\pi }{2}<\alpha <\frac{3\pi }{4}$.
Since $\alpha$ lies in the second quadrant, $\cos (\alpha )<0$. Thus, Statement II is true.

Next, we find the range of $\beta$.
Since $0<\frac{4}{9}<\frac{1}{2}$ and ${\cos }^{-1}(x)$ is a decreasing function, we have ${\cos }^{-1}\left(\frac{1}{2}\right)<{\cos }^{-1}\left(\frac{4}{9}\right)<{\cos }^{-1}(0)$.
$\Rightarrow \frac{\pi }{3}<{\cos }^{-1}\left(\frac{4}{9}\right)<\frac{\pi }{2}$
Multiplying by $3$, we get $\pi <\beta <\frac{3\pi }{2}$.
Thus, $\beta$ lies in the third quadrant.

Now, we determine the sign of $\cos (\alpha +\beta )$.
Adding the inequalities for $\alpha$ and $\beta$:
$\frac{\pi }{2}+\pi <\alpha +\beta <\frac{3\pi }{4}+\frac{3\pi }{2}$
$\Rightarrow \frac{3\pi }{2}<\alpha +\beta <\frac{9\pi }{4}$
The angle $(\alpha +\beta )$ lies in the interval $\left(\frac{3\pi }{2},\frac{9\pi }{4}\right)$, which covers the fourth quadrant and a part of the first quadrant. In both of these regions, the cosine function is strictly positive.
Therefore, $\cos (\alpha +\beta )>0$. Thus, Statement I is true.

Both Statement I and Statement II are true.

Answer: Both Statement I and Statement II are true