Question

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

• A. $\cot \theta$
• B. $\sec 2\theta$
• C. $\cot 2\theta$
• D. $\tan 2\theta$

Answer 1

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

$\frac{\sin 8\theta \cos \theta -\sin 6\theta \cos 3\theta }{\cos 2\theta \cos \theta -\sin 3\theta \sin 4\theta }=\frac{2\sin 8\theta \cos \theta -2\sin 6\theta \cos 3\theta }{2\cos 2\theta \cos \theta -2\sin 3\theta \sin 4\theta }=\frac{\sin 9\theta +\sin 7\theta -\sin 9\theta -\sin 3\theta }{\cos 3\theta +\cos \theta -\cos \theta +\cos 7\theta }=\frac{\sin 7\theta -\sin 3\theta }{\cos 3\theta +\cos 7\theta }=\frac{2\cos 5\theta \sin 2\theta }{2\cos 5\theta \cos 2\theta }=\tan 2\theta$