JEE Main 2026 — Mathematics Trigonometry

We are given $\alpha = 3\sin^{-1}\left(\dfrac{6}{11}\right)$α=3sin−1(116​) and $\beta = 3\cos^{-1}\left(\dfrac{4}{9}\right)$β=3cos−1(94​).

First, we find the range of $\alpha$α.

Since $\dfrac{1}{2} < \dfrac{6}{11} < \dfrac{1}{\sqrt{2}}$21​<116​<2​1​, we have $\sin^{-1}\left(\dfrac{1}{2}\right) < \sin^{-1}\left(\dfrac{6}{11}\right) < \sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right)$sin−1(21​)<sin−1(116​)<sin−1(2​1​).

$$\Rightarrow \dfrac{\pi}{6} < \sin^{-1}\left(\dfrac{6}{11}\right) < \dfrac{\pi}{4}$$⇒6π​<sin−1(116​)<4π​

Multiplying by $3$3, we get $\dfrac{\pi}{2} < \alpha < \dfrac{3\pi}{4}$2π​<α<43π​.

Since $\alpha$α lies in the second quadrant, $\cos(\alpha) < 0$cos(α)<0. Thus, Statement II is true.

Next, we find the range of $\beta$β.

Since $0 < \dfrac{4}{9} < \dfrac{1}{2}$0<94​<21​ and $\cos^{-1}(x)$cos−1(x) is a decreasing function, we have $\cos^{-1}\left(\dfrac{1}{2}\right) < \cos^{-1}\left(\dfrac{4}{9}\right) < \cos^{-1}(0)$cos−1(21​)<cos−1(94​)<cos−1(0).

$$\Rightarrow \dfrac{\pi}{3} < \cos^{-1}\left(\dfrac{4}{9}\right) < \dfrac{\pi}{2}$$⇒3π​<cos−1(94​)<2π​

Multiplying by $3$3, we get $\pi < \beta < \dfrac{3\pi}{2}$π<β<23π​.

Thus, $\beta$β lies in the third quadrant.

Now, we determine the sign of $\cos(\alpha+\beta)$cos(α+β).

Adding the inequalities for $\alpha$α and $\beta$β:

$$\dfrac{\pi}{2} + \pi < \alpha + \beta < \dfrac{3\pi}{4} + \dfrac{3\pi}{2}$$2π​+π<α+β<43π​+23π​

$$\Rightarrow \dfrac{3\pi}{2} < \alpha + \beta < \dfrac{9\pi}{4}$$⇒23π​<α+β<49π​

The angle $(\alpha + \beta)$(α+β) lies in the interval $\left(\dfrac{3\pi}{2}, \dfrac{9\pi}{4}\right)$(23π​,49π​), which covers the fourth quadrant and a part of the first quadrant. In both of these regions, the cosine function is strictly positive.

Therefore, $\cos(\alpha+\beta) > 0$cos(α+β)>0. Thus, Statement I is true.

Both Statement I and Statement II are true.

Answer: Both Statement I and Statement II are true