Question

The expression $\frac{\sin 8\theta \cos \theta -\sin 6\theta \cos 3\theta }{\cos 2\theta \cos \theta -\sin 3\theta \sin 4\theta }$ is equals -

• A. $\tan \theta$
• B. $\tan 2\theta$
• C. $\sin 2\theta$
• D. $\cos 2\theta$

Answer 1

The correct option is B $\tan 2\theta$

$\frac{\sin 8\theta \cos \theta -\sin 6\theta \cos 3\theta }{\cos 2\theta \cos \theta -\sin 3\theta \sin 4\theta }$
$=$$\frac{2\times (\sin 8\theta \cos \theta -\sin 6\theta \cos 3\theta )}{2\times (\cos 2\theta \cos \theta -\sin 3\theta \sin 4\theta )}$
$=$$\frac{2\sin 8\theta \cos \theta -2\sin 6\theta \cos 3\theta }{2\cos 2\theta \cos \theta -2\sin 3\theta \sin 4\theta }$
$=\frac{\sin (8\theta +\theta )+\sin (8\theta -\theta )-\sin (6\theta +3\theta )+\sin (6\theta -3\theta )}{\cos (2\theta +\theta )+\cos (2\theta -\theta )-\cos (3\theta -4\theta )-\cos (3\theta +4\theta )}$
$=\frac{sin7\theta -sin3\theta }{cos3\theta +\cos 7\theta }$
$=\frac{sin2\theta }{\cos 2\theta }=\tan 2\theta$