Question

If we have two lines $2\left( \sin a+\sin b \right)x-2\sin \left( a-b \right)y=3$ and $2\left( \cos a+\cos b \right)x+2\cos \left( a-b \right)y=5$ which are perpendicular, then $\sin 2a+\sin 2b=$.
(A). $\sin \left( a-b \right)-2\sin \left( a+b \right)=\sin \left( 2a \right)+\sin \left( 2b \right)$
(B). $\sin 2\left( a-b \right)-2\sin \left( a+b \right)=\sin \left( 2a \right)+\sin \left( 2b \right)$
(C). $2\sin \left( a-b \right)-\sin \left( a+b \right)=\sin \left( 2a \right)+\sin \left( 2b \right)$
(D). $\sin 2\left( a-b \right)-\sin \left( a+b \right)=\sin \left( 2a \right)+\sin \left( 2b \right)$

Answer 1

Hint: Here you have 2 line equations of form $ax+by=c$. Find the slope of the line by using geometry formulas. After getting both the slopes, use the condition of perpendicularly here. Slope of a line $ax+by+c=0$ is given by $\dfrac{-a}{b}$. Condition of perpendicularity of 2 lines of slope m, n is m.n = -1.

Complete step-by-step solution -
First line equation given in the question is:
$\Rightarrow 2\left( \sin a+\sin b \right)x-2\sin \left( a-b \right)y=3$
We know that the slope of the line $ax+by=c$ is $\dfrac{-a}{b}$.
By applying this here, assume the slope of this line is m.
By the above, we can say the value of m to be as:
$\Rightarrow m=\dfrac{-2\left( \sin a+\sin b \right)}{-2\sin \left( a-b \right)}$ ------ (1)
Second line equation given in question is written as follows:
$\Rightarrow 2\left( \cos a+\cos b \right)x+2\cos \left( a-b \right)y=5$
Let us assume the slope of the above line to be as n.
$\Rightarrow n=\dfrac{-2\left( \cos a+\cos b \right)}{2\cos \left( a-b \right)}$ ------ (2)
If 2 lines of slope m, n are perpendicular, we get:
$\Rightarrow m.n=-1$ ------ (3) (By condition of perpendicularity)
By multiplying equation (1) and equation (2), we get it as:
$\Rightarrow \dfrac{-2\left( \sin a+\sin b \right)}{-2\sin \left( a-b \right)}\times \dfrac{-2\left( \cos a+\cos b \right)}{2\cos \left( a-b \right)}=-1$
By cancelling common terms and do cross multiplication we get:
$\Rightarrow 2\left( \sin a+\sin b \right)\left( \cos a+\cos b \right)=2\sin \left( a-b \right)\cos \left( a-b \right)$
By simplify the left hand side, we get it as follows:
$\Rightarrow 2\sin a\cos a+2\sin b\cos b+2\sin a\cos b+2\cos a\sin b=2\sin \left( a-b \right)\cos \left( a-b \right)$
By knowledge of trigonometry, we know the formula as:
$\Rightarrow 2\sin x\cos x=\sin 2x$
By substituting this into our equation, we get it as:
$\Rightarrow \sin 2a+\sin 2b+2\sin a\cos b+2\cos a\sin b=\sin 2\left( a-b \right)$
By taking 2 common from last two terms, we get it as:
$\Rightarrow \sin 2a+\sin 2b+2\left( \sin a\cos b+\cos a\sin b \right)=\sin 2\left( a-b \right)$
By substituting $\left( \sin a\cos b+\cos a\sin b \right)$ as $\sin \left( a+b \right)$, we get it as:
$\Rightarrow \sin 2a+\sin 2b+2\sin \left( a+b \right)=\sin 2\left( a-b \right)$
By subtracting $\sin 2\left( a+b \right)$ on both sides, we get it as:
$\Rightarrow \sin 2\left( a-b \right)-\sin 2\left( a+b \right)=\sin 2a+\sin 2b$
Therefore option (b) is the correct answer for given conditions.

Note: Be careful while calculating slope from line equation, generally students forget the “-” sign and end up getting the wrong answer. So, the “-” sign in the slope is very important we left one 2 without cancelling because we want to apply $\sin 2x$ formula if you cancel then you get extra terms as $\dfrac{1}{2}$ on both sides of equation.