Kerala KEAM 5 June 2024 Solved Paper

We are given that $\sec(\alpha + \beta) = \frac{\sqrt{7}}{\sqrt{3}}$, and we are asked to find $\sin(\alpha + \beta) + \tan(\alpha + \beta)$.Step 1: From the given, $\sec(\alpha + \beta) = \frac{1}{\cos(\alpha + \beta)}$, so we can write: $$\cos(\alpha + \beta) = \frac{\sqrt{3}}{\sqrt{7}}.$$Step 2: Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, we can find $\sin(\alpha + \beta)$: $$\sin^2(\alpha + \beta) = 1 - \cos^2(\alpha + \beta) = 1 - \left( \frac{\sqrt{3}}{\sqrt{7}} \right)^2 = 1 - \frac{3}{7} = \frac{4}{7}.$$Thus, $$\sin(\alpha + \beta) = \frac{2}{\sqrt{7}}.$$Step 3: Now, we can calculate $\tan(\alpha + \beta)$ using the identity $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$: $$\tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{\frac{2}{\sqrt{7}}}{\frac{\sqrt{3}}{\sqrt{7}}} = \frac{2}{\sqrt{3}}.$$Step 4: Finally, we calculate the sum $\sin(\alpha + \beta) + \tan(\alpha + \beta)$: $$\sin(\alpha + \beta) + \tan(\alpha + \beta) = \frac{2}{\sqrt{7}} + \frac{2}{\sqrt{3}}.$$To combine these, we need a common denominator: $$\frac{2}{\sqrt{7}} + \frac{2}{\sqrt{3}} = \frac{2\sqrt{3} + 2\sqrt{7}}{\sqrt{21}} = \frac{2(\sqrt{3} + \sqrt{7})}{\sqrt{21}}.$$Thus, the correct answer is option (C). Quick Tip: When dealing with trigonometric identities, use $\sec(\theta) = \frac{1}{\cos(\theta)}$ to find $\cos(\theta)$, and use $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ to find $\tan(\theta)$.