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Question: How do you verify the identity \( \dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} + \dfrac{{\sin x - \s...

How do you verify the identity cos⁡x−cos⁡ysin⁡x+sin⁡y+sin⁡x−sin⁡ycos⁡x+cos⁡y=0\dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} + \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = 0 ?

Explanation

Solution

Hint : First we will evaluate the right-hand of the equation and then further the left-hand side of the equation. We will use the identity sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 . Then we will try to factorise and simplify the terms so that the left-hand side matches the right-hand side.

Complete step-by-step answer :
We will start off by solving the right-hand side of the equation. Here, we will be first making the denominator of the left-hand side the same.
Hence, the expression can be written as,
=cos⁡x−cos⁡ysin⁡x+sin⁡y+sin⁡x−sin⁡ycos⁡x+cos⁡y =(cos⁡x−cos⁡y)(cos⁡x+cos⁡y)+(sin⁡x−sin⁡y)(sin⁡x+sin⁡y)(sin⁡x+sin⁡y)(cos⁡x+cos⁡y)   = \dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} + \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} \\\ = \dfrac{{(\cos x - \cos y)(\cos x + \cos y) + (\sin x - \sin y)(\sin x + \sin y)}}{{(\sin x + \sin y)(\cos x + \cos y)}} \;
After we further simplify the expression it becomes,
=(cos⁡x−cos⁡y)(cos⁡x+cos⁡y)+(sin⁡x−sin⁡y)(sin⁡x+sin⁡y)(sin⁡x+sin⁡y)(cos⁡x+cos⁡y) =(cos⁡2x−cos⁡2y)+(sin⁡2x−sin⁡2y)(sin⁡x+sin⁡y)(cos⁡x+cos⁡y)  = \dfrac{{(\cos x - \cos y)(\cos x + \cos y) + (\sin x - \sin y)(\sin x + \sin y)}}{{(\sin x + \sin y)(\cos x + \cos y)}} \\\ = \dfrac{{({{\cos }^2}x - {{\cos }^2}y) + ({{\sin }^2}x - {{\sin }^2}y)}}{{(\sin x + \sin y)(\cos x + \cos y)}} \\\
Now we apply the trigonometric identity, sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 .
Hence, the expression becomes,
=(cos⁡2x−cos⁡2y)+(sin⁡2x−sin⁡2y)(sin⁡x+sin⁡y)(cos⁡x+cos⁡y) =1−cos⁡2y−sin⁡2y(sin⁡x+sin⁡y)(cos⁡x+cos⁡y)  = \dfrac{{({{\cos }^2}x - {{\cos }^2}y) + ({{\sin }^2}x - {{\sin }^2}y)}}{{(\sin x + \sin y)(\cos x + \cos y)}} \\\ = \dfrac{{1 - {{\cos }^2}y - {{\sin }^2}y}}{{(\sin x + \sin y)(\cos x + \cos y)}} \\\
Now we will again apply the trigonometric identity sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 but with some modification.
sin⁡2x+cos⁡2x=1 sin⁡2x=1−cos⁡2x   {\sin ^2}x + {\cos ^2}x = 1 \\\ {\sin ^2}x = 1 - {\cos ^2}x \;
Therefore, we will rewrite the expression.
=1−cos⁡2y−sin⁡2y(sin⁡x+sin⁡y)(cos⁡x+cos⁡y) =sin⁡2y−sin⁡2y(sin⁡x+sin⁡y)(cos⁡x+cos⁡y)   = \dfrac{{1 - {{\cos }^2}y - {{\sin }^2}y}}{{(\sin x + \sin y)(\cos x + \cos y)}} \\\ = \dfrac{{{{\sin }^2}y - {{\sin }^2}y}}{{(\sin x + \sin y)(\cos x + \cos y)}} \;
Now finally the expression will become,
=sin⁡2y−sin⁡2y(sin⁡x+sin⁡y)(cos⁡x+cos⁡y) =0(sin⁡x+sin⁡y)(cos⁡x+cos⁡y)  = \dfrac{{{{\sin }^2}y - {{\sin }^2}y}}{{(\sin x + \sin y)(\cos x + \cos y)}} \\\ = \dfrac{0}{{(\sin x + \sin y)(\cos x + \cos y)}} \\\
Now as we know that if the numerator is 00 the whole fraction becomes 00 .
Hence, the expression will become,
=0(sin⁡x+sin⁡y)(cos⁡x+cos⁡y) =0   = \dfrac{0}{{(\sin x + \sin y)(\cos x + \cos y)}} \\\ = 0 \;
Hence, LHS = 00
As, RHS = 00
Hence, we proved that cos⁡x−cos⁡ysin⁡x+sin⁡y+sin⁡x−sin⁡ycos⁡x+cos⁡y=0\dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} + \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = 0 .
So, the correct answer is “0”.

Note : While applying the double angle identities, first choose the identity according to the terms you have then choose the terms from the expression involving which you are using the double angle identities. While modifying any identity make sure that when you back trace the identity, you get the same original identity. LCM of the terms is one of the important factors during simplification.