Trigonometry

[Learningforcomp]

Hello, I am practicing trigonometry, and I've been given the following exercise

if (sinx - siny)/(cosx - cosy) = 2 and tanx = 1/3, then tan(y) is equal to:



I don't know how to solve this. I have tried to apply the sum and difference into product formulas but I got to an expression tan(x+y/2)=2 and don't know what to do.
Can you please help me?

[Baltuilhe]

Good night!

Given datum:
[tex]\tan x=\frac{1}{3}[/tex]

First, try to 'isolate':
[tex]\frac{\sin x-\sin y}{\cos x-\cos y}=2[/tex]
[tex]\frac{2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)}{-2\sin\left(\frac{x-y}{2}\right)\sin\left(\frac{x+y}{2}\right)}=2\\[/tex]
[tex]\frac{\cos\left(\frac{x+y}{2}\right)}{-\sin\left(\frac{x+y}{2}\right)}=2\\[/tex]
[tex]\tan\left(\frac{x+y}{2}\right)=\frac{-1}{2}[/tex]

Now, you have to solve for tan(x+y):
[tex]\tan(x+y)=\frac{2\tan\left(\frac{x+y}{2}\right)}{1-\tan^2\left(\frac{x+y}{2}\right)}[/tex]
[tex]\tan(x+y)=\frac{2\cdot\frac{-1}{2}}{1-\left(\frac{-1}{2}\right)^2}[/tex]
[tex]\tan(x+y)=\frac{-1}{1-\frac{1}{4}}=\frac{-1}{\frac{3}{4}}=\frac{-4}{3}[/tex]

Finally:
[tex]\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\cdot\tan y}[/tex]
[tex]\frac{-4}{3}=\frac{\frac{1}{3}+\tan y}{1-\frac{1}{3}\cdot\tan y}[/tex]
[tex]-4\cdot\left(1-\frac{\tan y}{3}\right)=3\cdot\left(\frac{1}{3}+\tan y\right)[/tex]
[tex]-4+\frac{4\tan y}{3}=1+3\tan y[/tex]

Multiplying by 3:
[tex]-12+4\tan y=3+9\tan y[/tex]
[tex]9\tan y-4\tan y=-12-3[/tex]
[tex]5\tan y=-15[/tex]
[tex]\tan y=\frac{-15}{5}=-3[/tex]

Hope to Have Helped (3H)