Question:
If \(\cos(\alpha + \beta) = 3/5\), \(\sin(\alpha - \beta) = 5/13\) and \(0 \lt \alpha, \beta \lt \frac{\pi}{4}\), then \(\tan(2\alpha)\) is equal to :
If \(\cos(\alpha + \beta) = 3/5\), \(\sin(\alpha - \beta) = 5/13\) and \(0 \lt \alpha, \beta \lt \frac{\pi}{4}\), then \(\tan(2\alpha)\) is equal to :
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Write \(2\alpha = (\alpha + \beta) + (\alpha - \beta)\) and use the tan addition formula after finding both tangents.
Updated On: Nov 5, 2026
- \(21/16\)
- \(63/52\)
- \(33/52\)
- \(63/16\)
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The Correct Option is D
Solution and Explanation
Since \(0 \lt \alpha + \beta \lt \frac{\pi}{2}\): \(\sin(\alpha + \beta) = \frac{4}{5}\), so \(\tan(\alpha + \beta) = \frac{4}{3}\).
Since \(-\frac{\pi}{4} \lt \alpha - \beta \lt \frac{\pi}{4}\) and \(\sin(\alpha - \beta) = \frac{5}{13} \gt 0\): \(\cos(\alpha - \beta) = \frac{12}{13}\), \(\tan(\alpha - \beta) = \frac{5}{12}\).
\(\tan 2\alpha = \tan[(\alpha + \beta) + (\alpha - \beta)] = \frac{\frac{4}{3} + \frac{5}{12}}{1 - \frac{4}{3}\cdot\frac{5}{12}} = \frac{\frac{21}{12}}{\frac{16}{36}} = \frac{7}{4}\times\frac{9}{4} = \frac{63}{16}\).
Answer: (4) 63/16
Since \(-\frac{\pi}{4} \lt \alpha - \beta \lt \frac{\pi}{4}\) and \(\sin(\alpha - \beta) = \frac{5}{13} \gt 0\): \(\cos(\alpha - \beta) = \frac{12}{13}\), \(\tan(\alpha - \beta) = \frac{5}{12}\).
\(\tan 2\alpha = \tan[(\alpha + \beta) + (\alpha - \beta)] = \frac{\frac{4}{3} + \frac{5}{12}}{1 - \frac{4}{3}\cdot\frac{5}{12}} = \frac{\frac{21}{12}}{\frac{16}{36}} = \frac{7}{4}\times\frac{9}{4} = \frac{63}{16}\).
Answer: (4) 63/16
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