Let … and …, where inverse trigonometric……

Single correct+4 / -1Answer matches NTA’s final key

Let $\alpha = 3\sin^{-1}\left(\frac{6}{11}\right)$ and $\beta = 3\cos^{-1}\left(\frac{4}{9}\right)$ , where inverse trigonometric functions take only the principal values.

Given below are two statements:

Statement I: $\cos(\alpha + \beta) > 0$ .

Statement II: $\cos(\alpha) < 0$ .

In the light of the above statements, choose the correct answer from the options given below:

Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true

Solution

Locate $\alpha$ . $\sin^{-1}\frac{6}{11}$ lies between $\sin^{-1}\frac12 = \frac{\pi}{6}$ and $\sin^{-1}\frac{1}{\sqrt2} = \frac{\pi}{4}$ (since $0.5 < 0.545 < 0.707$ ). So $\frac{\pi}{2} < \alpha < \frac{3\pi}{4}$ , which puts $\alpha$ in the second quadrant: $\cos\alpha < 0$ . Statement II is true.

Locate $\beta$ . $\cos^{-1}\frac49$ lies between $\cos^{-1}\frac12 = \frac{\pi}{3}$ and $\frac{\pi}{2}$ (since $0 < 0.444 < 0.5$ ). So $\pi < \beta < \frac{3\pi}{2}$ .

Combine more tightly. Numerically $\sin^{-1}\frac{6}{11} \approx 0.5769$ and $\cos^{-1}\frac49 \approx 1.1102$ , so $\alpha + \beta = 3(1.6871) \approx 5.061$ rad. That is between $\frac{3\pi}{2} \approx 4.712$ and $2\pi$ , the fourth quadrant, where cosine is positive. Statement I is true.

Answer: both statements are true.

Answer checked against the official answer key. The solution is our own working.

JEE MAIN 2026 · Trigonometry

Solution

Keep going