Question

$\frac{(\cos \alpha -\cos 3\alpha )(\sin 8\alpha +\sin 2\alpha )}{(\sin 5\alpha -\sin \alpha )(\cos 4\alpha -\cos 6\alpha )}=$

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $1$

$\Rightarrow \frac{-2\sin 2\alpha \sin \alpha \times 2\sin 5\alpha \cos 3\alpha }{2\cos 3\alpha \sin 2\alpha \times 2\sin \alpha \sin 5\alpha \times (-1)}$
$=1$

$\cos (A-B)=\cos A\cos B+\sin A\sin B$

$\sin (A-B)=\sin A\cos B-\cos A\sin B$

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

Hence, option 'A' is correct.