Question

Prove that
$\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \tan \dfrac{{x - y}}{2}$

Answer 1

Hint: Here, we have to prove $\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \tan \dfrac{{x - y}}{2}$. Take the LHS of the equation and try to prove LHS equal to RHS. For $\sin A - \sin B$, we have formula $\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$ and for $\cos A + \cos B$ we have formula $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$. Using these formulas, we will get LHS equal to RHS.

Complete step-by-step answer:
In this question, we are given a trigonometric equation and we need to prove that it is correct.
Given equation: $\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \tan \dfrac{{x - y}}{2}$- - - - - - - - - - - - (1)
Here, LHS$ = \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}}$ and RHS$ = \tan \dfrac{{x - y}}{2}$. So, we need to prove that LHS is equal to RHS.
For proving this, let us take the LHS part of the equation (1). Therefore, we get
$ \Rightarrow $LHS$ = \dfrac{{\sin x - \sin y}}{{\cos x + \cos y}}$- - - - - - - - - - - - (2)
Here, in numerator we have $\sin x - \sin y$. Now we know the formula that
$\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
And in denominator, we have $\cos x + \cos y$. Now, we know the formula that
 $\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
Therefore, putting these values in equation (2), we get
$ \Rightarrow $LHS$ = \dfrac{{2\cos \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right)}}{{2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)}}$- - - - - - - - - - - - - (3)
Here, 2 and $\cos \left( {\dfrac{{x + y}}{2}} \right)$ gets cancelled. Therefore, equation (3) will become
 $ \Rightarrow $LHS$ = \dfrac{{\sin \left( {\dfrac{{x - y}}{2}} \right)}}{{\cos \left( {\dfrac{{x - y}}{2}} \right)}}$- - - - - - - - - (4)
Now, we know that
$\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $
Therefore, putting this value in equation (4), we get
$ \Rightarrow $LHS$ = \tan \left( {\dfrac{{x - y}}{2}} \right)$
$ \Rightarrow LHS = RHS$
Hence, we have got LHS equal to RHS.
Therefore, we have proved that $\dfrac{{\sin x - \sin y}}{{\cos x + \cos y}} = \tan \dfrac{{x - y}}{2}$.

Note: This question is a simple formula based question. For solving trigonometric questions, always keep the important formulas and results in mind. Other important formulas are
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
$\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$