If sin(x+y) /sin(x-y) = a+b / a-b then what is tanx /tany equals to?

Question

It is asked from me in the nda examination. 

Answer 1

Recall the following two trigonometric identities:

\sin(x + y) = \sin(x)*\cos(y) + \cos(x)*\sin(y) (1.)

\sin(x - y) = \sin(x)*\cos(y) - \cos(x)*\sin(y) (2.)

Then since we are given that

\frac{\sin(x + y)}{\sin(x - y)}=\frac{a + b}{a - b},

we have the following two equations:

\sin(x + y) = a + b (3.)

\sin(x - y) = a - b (4.)

Adding (3.) and (4.) together gives us

\sin(x + y) + \sin(x - y) = (a + b) + (a - b)

= 2*a

⇒ \frac{\sin(x + y) + \sin(x - y)}{2} = a

⇒ a = \frac{\sin(x + y) + \sin(x - y)}{2}

Calculating (3.) - (4.) gives us

\sin(x + y) - \sin(x - y) = (a + b) - (a - b)

= a + b - a + b

= 2*b

⇒ \frac{\sin(x + y) - \sin(x - y)}{2} = b

⇒ b = \frac{\sin(x + y) - \sin(x - y)}{2}

Substituting in (1.) and (2.) into the formula for a then gives us

a = \frac{[\sin(x)*\cos(y) + \cos(x)*\sin(y)] + [\sin(x)*\cos(y) - \cos(x)*\sin(y)]}{2}

= \frac{2*sin(x)*cos(y)}{2}

= sin(x)*cos(y)

Substituting in (1.) and (2.) into the formula for b then gives us

b = \frac{[\sin(x)*\cos(y) + \cos(x)*\sin(y)] - [\sin(x)*\cos(y) - \cos(x)*\sin(y)]}{2}

= \frac{\sin(x)*\cos(y) + \cos(x)*\sin(y) - \sin(x)*\cos(y) + \cos(x)*\sin(y)}{2}

= \frac{2*\cos(x)*\sin(y)}{2}

= \cos(x)*\sin(y)

Finally, we observe that

\frac{\tan(x)}{\tan(y)} = \frac{\frac{\sin(x)}{\cos(x)}}{\frac{\sin(y)}{\cos(y)}}

= [\frac{\sin(x)}{\cos(x)}]*[\frac{\cos(y)}{\sin(y)}]

= \frac{\sin(x)*\cos(y)}{\cos(x)*\sin(y)}

= \frac{a}{b}