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.