Question

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

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

Answer 1

Answer: (B)

$\sin 2(a-b)-2\sin (a+b)=\sin \left(2a\right)+\sin \left(2b\right)$

Given equations are

$2(\sin a+\sin b)x-2\sin (a-b)y=3\to \left(1\right)$

and $2(\cos a+\cos b)x+2\cos (a-b)y=5\to \left(2\right)$

Let the slope of the equation $\left(1\right)$ be $m_1$

and equation $\left(2\right)$ be $m_2$

$m_1=\frac{\sin a+\sin b}{\sin (a-b)}\to \left(3\right)$

$m_2=-\frac{\cos a+\cos b}{\cos (a-b)}\to \left(4\right)$

Since both the lines are mutually perpendicular $\left(m_1\right)\left(m_2\right)=-1\to \left(5\right)$

From $\left(3\right)$, $\left(4\right)$ and $\left(5\right)$

Using the identities, $\cos 2\alpha ={\cos }^2\left(\alpha \right)-{\sin }^2\left(\alpha \right)$

$\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$

We get $m_1{m}_2=-1$

$\Rightarrow \frac{\sin a\cos a+\sin a\cos b+\sin b\cos a+\sin b\cos b}{\sin (a-b)\cos (a-b)}=1$

$\Rightarrow \frac{\sin (a+b)+\frac{1}{2}[\sin 2b+\sin 2a]}{\sin (a-b)\cos (a-b)}=1$

$\Rightarrow \sin \left(2a\right)+\sin \left(2b\right)=2\left[\frac{1}{2}\sin \right(2a-2b)-\sin (a+b\left)\right]$

$=\sin 2(a-b)-2\sin (a+b)$

We get, $\sin 2a+\sin 2b=\sin 2(a-b)-2\sin (a+b)$