<p><strong>Ex. 62:</strong> If in triangle ABC, $\tan A + \tan B + \tan C = 6$ and $\tan A \tan B = 2$, then $\sin^2 A : \sin^2 B : \sin^2 C$ is</p>
Step-by-Step Solution
Key Concept: Use the identity $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ for triangle ABC to find individual angles, then convert tangent to sine.
<p><strong>Step 1:</strong> Given $\tan A + \tan B + \tan C = 6$ and $\tan A \tan B = 2$</p><p><strong>Step 2:</strong> Since $A + B + C = \pi$, we have $\tan A + \tan B + \tan C = \tan A \tan B \tan C$</p><p>Therefore, $\tan A \tan B \tan C = 6$</p><p><strong>Step 3:</strong> From $\tan A \tan B = 2$, we get $2 \tan C = 6$, so $\tan C = 3$</p><p><strong>Step 4:</strong> $\sin^2 C = \frac{\tan^2 C}{1 + \tan^2 C} = \frac{9}{1+9} = \frac{9}{10}$</p><p><strong>Step 5:</strong> From $\tan A + \tan B = 3$ and $\tan A \tan B = 2$, we get $\tan A - \tan B = \pm\sqrt{(\tan A + \tan B)^2 - 4\tan A \tan B} = \pm\sqrt{9-8} = \pm 1$</p><p>Thus $\tan A = 2, 1$ and $\tan B = 1, 2$</p><p><strong>Step 6:</strong> $\sin^2 A = \frac{4}{5}$ and $\sin^2 B = \frac{1}{5}$, or vice versa</p><p>∴ Answer is (b) $8:5:9$</p>
Correct Answer: B