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Question
The value of \[\frac{\sin 5 \alpha - \sin 3\alpha}{\cos 5 \alpha + 2 \cos 4\alpha + \cos 3\alpha} =\]
Options
- \[\cot \alpha/2\]
- \[\cot \alpha\]
- \[\tan \alpha/2\]
None of these
Solution
\[\frac{\sin5\alpha - \sin3\alpha}{\cos5\alpha + 2\cos4\alpha + \cos3\alpha} = \frac{\sin5\alpha - \sin3\alpha}{\cos5\alpha + \cos3\alpha + 2\cos4\alpha}\]
\[ = \frac{2\sin\alpha\cos4\alpha}{2\cos4\alpha\cos\alpha + 2\cos4\alpha}\]
\[ = \frac{2\sin\alpha\cos4\alpha}{2\cos4\alpha\left( \cos\alpha + 1 \right)}\]
\[ = \frac{\sin\alpha}{\cos\alpha + 1}\]
\[ = \frac{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{\cos^2 \frac{\alpha}{2} - \sin^2 \frac{\alpha}{2} + \sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}}\]
\[ = \frac{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2 \cos^2 \frac{\alpha}{2}}\]
\[ = \frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}\]
\[ = \tan\frac{\alpha}{2}\]
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