Question

The value of $\frac{\sin 5\alpha -\sin 3\alpha }{\cos 5\alpha +2\cos 4\alpha +\cos 3\alpha }=$
(a) $\cot \alpha /2$
(b) $\cot \alpha$
(c) $\tan \alpha /2$
(d) None of these

Answer 1

(c) $\tan \alpha /2$

$$\begin{gathered} \frac{\sin 5\alpha -\sin 3\alpha }{\cos 5\alpha +2\cos 4\alpha +\cos 3\alpha }=\frac{\sin 5\alpha -\sin 3\alpha }{\cos 5\alpha +\cos 3\alpha +2\cos 4\alpha } \\ =\frac{2\sin \alpha \cos 4\alpha }{2\cos 4\alpha \cos \alpha +2\cos 4\alpha } \\ =\frac{2\sin \alpha \cos 4\alpha }{2\cos 4\alpha \left(\cos \alpha +1\right)} \\ =\frac{\sin \alpha }{\cos \alpha +1} \\ =\frac{2\sin {\displaystyle \frac{\alpha }{2}}\cos {\displaystyle \frac{\alpha }{2}}}{{\cos }^2{\displaystyle \frac{\alpha }{2}}-{\sin }^2{\displaystyle \frac{\alpha }{2}}+{\sin }^2{\displaystyle \frac{\alpha }{2}}+{\cos }^2{\displaystyle \frac{\alpha }{2}}} \\ =\frac{2\sin {\displaystyle \frac{\alpha }{2}}\cos {\displaystyle \frac{\alpha }{2}}}{2{\cos }^2{\displaystyle \frac{\alpha }{2}}} \\ =\frac{\sin {\displaystyle \frac{\alpha }{2}}}{\cos {\displaystyle \frac{\alpha }{2}}} \\ =\tan \frac{\alpha }{2} \end{gathered}$$