Question

If $ \sin x + \sin y = \dfrac{3}{4} $ and $ \sin x - \sin y = \dfrac{2}{5} $ , then $ \dfrac{{\tan \left( {\dfrac{{x - y}}{2}} \right)}}{{\tan \left( {\dfrac{{x + y}}{2}} \right)}} = ? $
(A) $ \dfrac{{15}}{8} $
(B) $ \dfrac{8}{{15}} $
(C) $ \dfrac{3}{{10}} $
(D) $ \dfrac{{10}}{3} $

Answer 1

Hint: Use the transformation formulae of $ \sin x \pm \sin y $ . Then divide both the equations to write then in terms of tan. Then substitute the given values in the equation you get to solve the question. Be careful about the division of the fractions while solving it.

Complete step-by-step answer:
By using the transformation formulae, we can write
 $ \sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right) $ . . . (1)
and
 $ \sin x - \sin y = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right) $ . . . (2)
By dividing equation (2) by equation (1), we get
\[\dfrac{{\sin x - \sin y}}{{\sin x + \sin y}} = \dfrac{{2\cos \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{x - y}}{2}} \right)}}{{2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)}}\]
By cancelling the common terms and rearranging it, we can write
\[\dfrac{{\sin x - \sin y}}{{\sin x + \sin y}} = \dfrac{{\sin \left( {\dfrac{{x - y}}{2}} \right)}}{{\cos \left( {\dfrac{{x - y}}{2}} \right)}} \times \dfrac{1}{{\dfrac{{\sin \left( {\dfrac{{x + y}}{2}} \right)}}{{\cos \left( {\dfrac{{x + y}}{2}} \right)}}}}\] \[\left( {\because x = \dfrac{1}{{\dfrac{1}{x}}}} \right)\]
We know that,
 $ \dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $
Using this property, we can write the above equation as
\[\dfrac{{\sin x - \sin y}}{{\sin x + \sin y}} = \dfrac{{\tan \left( {\dfrac{{x - y}}{2}} \right)}}{{\tan \left( {\dfrac{{x + y}}{2}} \right)}}\]
Now, it is given in the question that,
 $ \sin x - \sin y = \dfrac{2}{5} $
 $ \sin x + \sin y = \dfrac{3}{4} $
Substituting the given values in the above equation. We get
\[\dfrac{{\dfrac{2}{5}}}{{\dfrac{3}{4}}} = \dfrac{{\tan \left( {\dfrac{{x - y}}{2}} \right)}}{{\tan \left( {\dfrac{{x + y}}{2}} \right)}}\]
Rearranging it we can write
\[\dfrac{{\tan \left( {\dfrac{{x - y}}{2}} \right)}}{{\tan \left( {\dfrac{{x + y}}{2}} \right)}} = \dfrac{{2 \times 4}}{{5 \times 3}}\] $ \left( {\because \dfrac{{\dfrac{a}{b}}}{{\dfrac{p}{q}}} = \dfrac{a}{b} \times \dfrac{q}{p}} \right) $
 $ \Rightarrow \dfrac{{\tan \left( {\dfrac{{x - y}}{2}} \right)}}{{\tan \left( {\dfrac{{x + y}}{2}} \right)}} = \dfrac{8}{{15}} $
Therefore, from the above explanation, the correct answer is, option (B) $ \dfrac{8}{{15}} $
So, the correct answer is “Option B”.

Note: The key point of this question was to remember the transformation formula. It is not that you won’t be able to solve this question without the transformation formula. The alternative would be to right tan in terms of sin and cos and then expand $ \sin (x \pm y) $ and $ \cos (x \pm y) $ . The simplify the equation formed until it looks at the form that is given in the question. And then substitute the values in it to get the answer. This would be a long method. Knowing transformation formulae made it look short and easy.