Truth Value of Statements Involving Inverse Trigonometric Functions and Cosine

To determine the truth value of the given statements, let us analyze $\alpha$ and $\beta$ step-by-step using principal values.

Step 1: Evaluation of Statement II

We are given: $\alpha = 3 \sin^{-1} \left(\frac{6}{11}\right)$

Let $x = \sin^{-1} \left(\frac{6}{11}\right)$. Since $\frac{1}{2} < \frac{6}{11} < \frac{\sqrt{2}}{2}$, we have: $\sin\left(\frac{\pi}{6}\right) < \sin(x) < \sin\left(\frac{\pi}{4}\right)$

Since $\sin \theta$ is an increasing function on $\left[0, \frac{\pi}{2}\right]$, it follows that: $\frac{\pi}{6} < x < \frac{\pi}{4}$

Multiplying the inequality by $3$: $3 \times \frac{\pi}{6} < 3x < 3 \times \frac{\pi}{4} \implies \frac{\pi}{2} < \alpha < \frac{3\pi}{4}$

Since $\alpha$ lies in the second quadrant, its cosine value is negative: $\cos(\alpha) < 0$

Thus, Statement II is TRUE.

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Step 2: Evaluation of Statement I

We are given: $\beta = 3 \cos^{-1} \left(\frac{4}{9}\right)$

Let $y = \cos^{-1} \left(\frac{4}{9}\right)$, so $x + y = \sin^{-1} \left(\frac{6}{11}\right) + \cos^{-1} \left(\frac{4}{9}\right)$.

From the definitions of $x$ and $y$ in the first quadrant:

• $\sin x = \frac{6}{11} \implies \cos x = \sqrt{1 - \left(\frac{6}{11}\right)^2} = \frac{\sqrt{85}}{11}$
• $\cos y = \frac{4}{9} \implies \sin y = \sqrt{1 - \left(\frac{4}{9}\right)^2} = \frac{\sqrt{65}}{9}$

Using the cosine addition formula: $\cos(x + y) = \cos x \cos y - \sin x \sin y$ $\cos(x + y) = \left(\frac{\sqrt{85}}{11}\right)\left(\frac{4}{9}\right) - \left(\frac{6}{11}\right)\left(\frac{\sqrt{65}}{9}\right) = \frac{4\sqrt{85} - 6\sqrt{65}}{99}$

Comparing $4\sqrt{85}$ and $6\sqrt{65}$: $(4\sqrt{85})^2 = 16 \times 85 = 1360$ $(6\sqrt{65})^2 = 36 \times 65 = 2340$

Since $1360 < 2340$, $4\sqrt{85} < 6\sqrt{65}$, which means: $\cos(x + y) < 0 \implies x + y > \frac{\pi}{2}$

Now, let us compare $\cos(x + y)$ with $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$: $\cos(x + y) - \left(-\frac{1}{2}\right) = \frac{4\sqrt{85} - 6\sqrt{65}}{99} + \frac{1}{2} = \frac{8\sqrt{85} - 12\sqrt{65} + 99}{198}$

Approximating the square roots:

• $8\sqrt{85} > 8 \times 9.2 = 73.6$
• $12\sqrt{65} < 12 \times 8.1 = 97.2$

Thus: $8\sqrt{85} - 12\sqrt{65} + 99 > 73.6 - 97.2 + 99 = 75.4 > 0$

Therefore, $\cos(x + y) > -\frac{1}{2} = \cos\left(\frac{2\pi}{3}\right)$.

Since $\cos\theta$ is strictly decreasing on $[0, \pi]$: $\frac{\pi}{2} < x + y < \frac{2\pi}{3}$

Multiplying through by $3$: $\frac{3\pi}{2} < 3(x + y) < 2\pi \implies \frac{3\pi}{2} < \alpha + \beta < 2\pi$

Since $\alpha + \beta$ lies strictly in the fourth quadrant, its cosine value is positive: $\cos(\alpha + \beta) > 0$

Thus, Statement I is TRUE.

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Conclusion

Both Statement I and Statement II are true.

Correct Option: A