Question

If the lines $2(\sin \,a+\sin \,b) x-2 \,\sin\, (a-b) y=3$ and $2(\cos\, a$ $+\cos \,b) x+2 \,\cos (a-b) y=5$ are perpendicular, then $\sin \,2 a+$ $\sin \,2 b$ is equal to

• A. $\sin (a-b)-2 \sin (a+b)$
• B. $\sin\, 2(a-b)-2\, \sin (a+b)$
• C. $2 \,\sin (a-b)-\sin (a+b)$
• D. $\sin\, 2(a-b)-\sin (a+b)$

Answer 1

Answer: (B)

We have $\frac{\sin \,a+\sin \,b}{\sin (a-b)} \times-\frac{\cos \,a+\cos \,b}{\cos (a-b)}=-1$
$\Rightarrow \sin \,2 a+\sin\, 2 b+2 \,\sin (a+b)=\sin \,2(a-b)$
$\therefore \sin\, 2 a+\sin \,2 b=\sin \,2(a-b)-2\, \sin (a+b)$