Question

Prove tanx-tany/tanx+tany = sin(x-y)/sin(x+y)

Answer 1

@study100

Answer 2

Ok you need to manipulate this a little bit

Answer 3

\[\frac{ \tan(x) - \tan(y) }{ \tan(x) + \tan(y) }= \frac{ \sin(x-y) }{ \sin(x+y) }\]
Let's work with the left side.

Answer 4

\[\large \frac{ (\frac{ \sin(x) }{ \cos(x) })-(\frac{ \sin(y) }{ \cos(y) } )}{ (\frac{ \sin(x) }{ \cos(x) })+(\frac{ \sin(y) }{ \cos(y) }) }\]

Then I will cross multiply the numerator, then cross multiply the denominator.

Answer 5

\[\frac{ \sin(x)\cos(y) - \cos(x)\sin(y) }{ \cos(x)\sin(y)+\sin(x)\cos(y) }\]

Answer 6

Now let's just look at the denominator

\[-\cos(x)\sin(y)+\cos(y)\sin(x)\]

You know based on the Sum and Difference identity, sin (u+v)= sin u cos v +- cos u sin v. Therefore,
\[-\cos(x)\sin(y)+\cos(y)\sin(x) = \sin(x)\cos(y) - \cos(x)\sin(y) \]
\[= \sin(x-y)\]

Answer 7

oh sorry, that was the numerator (top)

now it will be similar for the bottom

Answer 8

\[\cos(x)\sin(y)+\cos(y)\sin(x) = \sin(x)\cos(y) + \cos(x)\sin(y)\]
= sin(x+y)

Answer 9

Now that we have simplified, the top and bottom we get
\[\frac{ \sin(x-y) }{ \sin(x+y) }\]

which is the same as the right side.

so this identiy is proven to be true.

Answer 10

Thanks so much for your help