Question

How do you verify the identity\[\dfrac{\sin x+\sin y}{\cos x-\cos y}=-\cot \left( \dfrac{x-y}{2} \right)\]?

Answer 1

Hint: For the given question we are given a trigonometric equation to prove LHS=RHS. First step for doing this problem we have to take any of the parts of either LHS or RHS and then we should prove the other part. By applying the identities \[\begin{align}
  & \sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right) \\
 & \cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right) \\
\end{align}\] our problem will come to end. By solving the remaining equation we can prove our equation.

Complete step by step answer:
For the given question we are given to verify the identity\[\dfrac{\sin x+\sin y}{\cos x-\cos y}=-\cot \left( \dfrac{x-y}{2} \right)\].
Now for verifying the above equation let us derive the RHS (right and side) part by the LHS (left hand side).
Let us consider the given equation as equation (1).
\[\dfrac{\sin x+\sin y}{\cos x-\cos y}=-\cot \left( \dfrac{x-y}{2} \right)............\left( 1 \right)\]
Now let us take the LHS part and derive, for that let us assume the equation as ‘S’.
\[\Rightarrow S=\dfrac{\sin x+\sin y}{\cos x-\cos y}=-\cot \left( \dfrac{x-y}{2} \right)\]
Let us consider the above equation as equation (2).
\[S=\dfrac{\sin x+\sin y}{\cos x-\cos y}=-\cot \left( \dfrac{x-y}{2} \right).........\left( 2 \right)\]
Taking LHS,
\[\Rightarrow S=\dfrac{\sin x+\sin y}{\cos x-\cos y}\]
Let us consider this as equation (3).
\[S=\dfrac{\sin x+\sin y}{\cos x-\cos y}...............\left( 3 \right)\]
As we know the identities\[\begin{align}
  & \sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right) \\
 & \cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right) \\
\end{align}\]. Now let us consider these as identity ($I_{1}$) and identity ($I_{2}$).
\[\begin{align}
  & \sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right).............\left( I_1 \right) \\
 & \cos A-\cos B=2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right).............\left( I_2 \right) \\
\end{align}\]
By applying the identity ($I_{1}$) to the numerator of equation (3) as well as Identity ($I_{2}$) to the denominator of equation (3), we get
\[\Rightarrow S=\dfrac{2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)}{2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)}\]
Let us consider the above equation as equation (3).
\[\Rightarrow S=\dfrac{2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)}{2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{B-A}{2} \right)}.........\left( 3 \right)\]
By simplifying the equation (3), we get
\[\Rightarrow S=\dfrac{\cos \left( \dfrac{A-B}{2} \right)}{\sin \left( \dfrac{B-A}{2} \right)}\]
Let us consider the above equation as equation (4).
\[\Rightarrow S=\dfrac{\cos \left( \dfrac{A-B}{2} \right)}{\sin \left( \dfrac{B-A}{2} \right)}...........\left( 4 \right)\]
As we know the identity\[\sin \left( -A \right)=-\sin \left( A \right)\]. Let us consider it as identity ($I_{3}$).
By observing the denominator of the equation we can see a term i.e.\[\sin \left( \dfrac{B-A}{2} \right)\]. Let us apply the identity ($I_{3}$) to the equation (4), we get
\[\Rightarrow S=-\dfrac{\cos \left( \dfrac{A-B}{2} \right)}{\sin \left( \dfrac{A-B}{2} \right)}\]
As we know the formula\[\dfrac{\cos a}{\sin a}=\cot a\]. By applying the formula to the above equation, we get
\[\Rightarrow S=-\cot \left( \dfrac{A-B}{2} \right)\]
Let us consider the above equation as equation (5).
\[\Rightarrow S=-\cot \left( \dfrac{A-B}{2} \right).........\left( 5 \right)\]
Therefore from equation (2) and equation (5) we can say that,
\[S=\dfrac{\sin x+\sin y}{\cos x-\cos y}=-\cot \left( \dfrac{x-y}{2} \right)\]
LHS=RHS
Hence, proved.

Note: We can solve the above problem by taking RHS as a reference also. Students should be aware of all trigonometric identities for solving the above problem. If students apply the \[\cos A-\cos B\] as \[2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{B-A}{2} \right)\] then whole problem will be wrong, so students should learn identities perfectly.