Help verifying trig identities

**sam1219** · May 4th 2011, 10:18 AM

I'm having a little bit of trouble verifying 2 trig identities. One of these may not even be an identity but I don't know which one since I'm having trouble with both.
The first one is:

sin(x+y)/sin(x-y) = 1-tanxtany/1+tanxtany

And the second one is:

(sinx/1-cotx) - (cosx/tanx-1) = sinx+cosx

I get to a certain point on both of these and can't figure out where to go from there, any help in how to verify these would be GREATLY appreciated!!

Thanks!

**abhishekkgp** · May 4th 2011, 10:24 AM

Originally Posted by sam1219

I'm having a little bit of trouble verifying 2 trig identities. One of these may not even be an identity but I don't know which one since I'm having trouble with both.
The first one is:

sin(x+y)/sin(x-y) = 1-tanxtany/1+tanxtany

This is wrong. you should take reciprocal of the RHS to fix this one

And the second one is:

(sinx/1-cotx) - (cosx/tanx-1) = sinx+cosx
please bracket it properly so that i can give you a definitive answer.

I get to a certain point on both of these and can't figure out where to go from there, any help in how to verify these would be GREATLY appreciated!!

Thanks!

did this help?

**sam1219** · May 4th 2011, 11:53 AM

So are you saying the first one is not an identity? That is the problem that was given to me and I'm asked to verify it, so by wrong do you mean it's not an identity?

and I'm not sure how you want me to bracket the second one properly, but here's another try at what it looks like

Sinx - cosx = sinx + cosx
1-cotx tanx-1

**Soroban** · May 4th 2011, 01:07 PM

Hello, sam1219!

$\frac{\sin(x+y)}{\sin(x-y)} \:=\: \frac{1-\tan x\tan y}{1+\tan x\tan y}$

This one is not an identity . . .

$\frac{\sin(x+y)}{\sin(x-y)} \:=\: \frac{\sin x\cos y + \cos x\sin y}{\sin x\cos y - \cos x\sin y}$

$\text{Divide numerator and denominator by }\sin x\cos y\!:$

. . $\frac{\dfrac{\sin x\cos y}{\sin x\cos y} + \dfrac{\cos x\sin y}{\sin x\cos y}} {\dfrac{\sin x\cos y}{\sin x\cos y} - \dfrac{\cos x\sin y}{\sin x\cos y}} \;=\;\frac{1 + \cot x\tan y}{1 - \cot x\tan y}$

. . $=\;\frac{1 + \dfrac{\tan y}{\tan x}}{1 - \dfrac{\tan y}{\tan x}} \;=\; \frac{\tan x + \tan y}{\tan x - \tan y}$

$\frac{\sin x}{1-\cot x} - \frac{\cos x}{\tan x-1} \:=\: \sin x+\cos x$

$\text{We have: }\;\frac{\sin x}{1 - \frac{\cos x}{\sin x}} - \frac{\cos x}{\frac{\sin x}{\cos x} - 1}} \;\;=\;\;\frac{\sin x}{\sin x}\cdot\frac{\sin x}{1 - \frac{\cos x}{\sin x}} \:-\:\frac{\cos x}{\cos x}\cdot\frac{\cos x}{\frac{\sin x}{\cos x} - 1}$

. . $=\;\;\frac{\sin^2\!x}{\sin x- \cos x} - \frac{\cos^2\!x}{\sin x - \cos x} \;\;=\;\; \frac{\sin^2\!x - \cos^2\!x}{\sin x - \cos x}$

. . $=\;\;\frac{(\sin x - \cos x)(\sin x + \cos x)}{\sin x - \cos x} \;\;=\;\;\sin x + \cos x$

**sam1219** · May 4th 2011, 06:27 PM

Thank you so much! You are awesome!!

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