Question

$\frac{\cos 8A\cos 5A-\cos 12A\cos 9A}{\sin 8A\cos 5A+\cos 12\sin 9A}=\tan 4A$.

• A. True
• B. False

Answer 1

The correct option is A True

$\cos 8A\cos 5A=\frac{1}{2}(\cos 13A+\cos 3A)$

$\cos 12A\cos 9A=\frac{1}{2}(\cos 21A+\cos 3A)$

$\sin 8A\cos 5A=\frac{1}{2}(\sin 13A+\sin 3A)$

$\cos 12A\sin 9A=\frac{1}{2}(\sin 21A-\sin 3A)$

$\frac{\cos 8A\cos 5A-\cos 12A\cos 9A}{\sin 8A\cos 5A+\cos 12A\sin 9A}=\frac{\frac{1}{2}(\cos 13A+\cos 3A-\cos 21A-\cos 3A)}{\frac{1}{2}(\sin 13A+\sin 3A+\sin 21A-\sin 3A)}$

$=\frac{\cos 13A-\cos 21A}{\sin 13A+\sin 21A}$

$=\frac{2\sin 17A\sin 4A}{2\sin 17A\cos 4A}$

$=\tan 4A$

So the given relation is $\text{True}$