Question

How do you verify the following identity
\[\dfrac{{\cos x - \cos y}}{{\sin x + \sin y}}\] + \[\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}}\] = 0

Answer 1

Hint: Here, you are given an equation which includes the trigonometric ratios sine and cosine and are asked to verify the identity. What you need to do is to use the trigonometric relation between sine and cosine of an angle and try to simplify the left-hand side. By simplifying what is meant is you need to bring down the left-hand side and make it equal to the right-hand side by giving the term a different form, that is, you need to rearrange the terms by cross multiplying the terms.

Complete step by step solution:
Let us consider the left-hand side. It consists of two terms, one is \[\dfrac{{\cos x - \cos y}}{{\sin x + \sin y}}\] and the other is \[\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}}\] . You need to add these two terms. As you can see that both the terms are fractions. In order to add two fractions, you need to have the denominators of both the fractions equal. What you can do is multiply and divide the denominator of first term to the second term and that of second term to the first term. We will get the first term as \[\dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} = \dfrac{{\cos x + \cos y}}{{\cos x + \cos y}}\left( {\dfrac{{\cos x - \cos y}}{{\sin x + \sin y}}} \right)\] and the second term as \[\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \dfrac{{\sin x + \sin y}}{{\sin x + \sin y}}\left( {\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}}} \right)\] . You can see that the denominators are equal now. Let us multiply and expand the terms.
The first term will be
 \[
\Rightarrow \dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} = \dfrac{{\cos x + \cos y}}{{\cos x + \cos y}}\left( {\dfrac{{\cos x - \cos y}}{{\sin x + \sin y}}} \right) \\
\Rightarrow \dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} = \dfrac{{\left( {\cos x + \cos y} \right)\left( {\cos x - \cos y} \right)}}{{\left( {\cos x + \cos y} \right)\left( {\sin x + \sin y} \right)}} \\
\Rightarrow \dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} = \dfrac{{{{\cos }^2}x - \cos x\cos y + \cos x\cos y - {{\cos }^2}y}}{{\left( {\cos x + \cos y} \right)\left( {\sin x + \sin y} \right)}} \\
\Rightarrow \dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} = \dfrac{{{{\cos }^2}x - {{\cos }^2}y}}{{\left( {\cos x + \cos y} \right)\left( {\sin x + \sin y} \right)}} \;
 \]
The second term will be
 \[
\Rightarrow \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \dfrac{{\sin x + \sin y}}{{\sin x + \sin y}}\left( {\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}}} \right) \\
\Rightarrow \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \dfrac{{\left( {\sin x + \sin y} \right)\left( {\sin x - \sin y} \right)}}{{\left( {\sin x + \sin y} \right)\left( {\cos x + \cos y} \right)}} \\
\Rightarrow \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \dfrac{{{{\sin }^2}x - \sin x\sin y + \sin x\sin y - {{\sin }^2}y}}{{\left( {\sin x + \sin y} \right)\left( {\cos x + \cos y} \right)}} \\
\Rightarrow \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \dfrac{{{{\sin }^2}x - {{\sin }^2}y}}{{\left( {\sin x + \sin y} \right)\left( {\cos x + \cos y} \right)}} \;
 \]
Let us add the two terms,
 \[
\Rightarrow \dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} + \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \dfrac{{{{\cos }^2}x - {{\cos }^2}y}}{{\left( {\cos x + \cos y} \right)\left( {\sin x + \sin y} \right)}} + \dfrac{{{{\sin }^2}x - {{\sin }^2}y}}{{\left( {\sin x + \sin y} \right)\left( {\cos x + \cos y} \right)}} \\
\Rightarrow \dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} + \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \dfrac{{{{\cos }^2}x - {{\cos }^2}y + {{\sin }^2}x - {{\sin }^2}y}}{{\left( {\cos x + \cos y} \right)\left( {\sin x + \sin y} \right)}} \;
 \]
Now, we have a trigonometric identity which states that the sum of the squares of sine and cosine of an angle is equal to one. Mathematically, we have, $ {\sin ^2}x + {\cos ^2}x = 1 $ . You can see that if we rearrange the above obtained expression, we can use this identity to simplify it.
 \[
\Rightarrow \dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} + \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \dfrac{{{{\cos }^2}x - {{\cos }^2}y + {{\sin }^2}x - {{\sin }^2}y}}{{\left( {\cos x + \cos y} \right)\left( {\sin x + \sin y} \right)}} \\
\Rightarrow \dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} + \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \dfrac{{\left( {{{\cos }^2}x + {{\sin }^2}x} \right) - \left( {{{\cos }^2}y + {{\sin }^2}y} \right)}}{{\left( {\cos x + \cos y} \right)\left( {\sin x + \sin y} \right)}} \\
\Rightarrow \dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} + \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \dfrac{{\left( 1 \right) - \left( 1 \right)}}{{\left( {\cos x + \cos y} \right)\left( {\sin x + \sin y} \right)}} = 0 \;
 \]
Hence proved \[\dfrac{{\cos x - \cos y}}{{\sin x + \sin y}} + \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = 0\] , identity verified.

Note: Here, we have discussed in detail how we reached the final expression so that the whole left-hand side is proved to be zero. You need to memorize the trigonometric identities and properties such as the relation between the sine and cosine of an angle, that is $ {\sin ^2}x + {\cos ^2}x = 1 $ , so that you can easily or at least get a head start to initiate how to solve the question. You have to imagine which property can be used in order to simplify the equation.