# Solving a trig identity

• Nov 20th 2010, 08:32 PM
youngb11
Solving a trig identity
LS=(tanx-tany)/(cotx-coty)

LS=-tanxtany

If possible try to avoid use of the tan compound formula.

Thanks!
• Nov 20th 2010, 08:33 PM
dwsmith
What are you trying to do?
• Nov 21st 2010, 12:08 AM
red_dog
$\displaystyle \dfrac{\tan x-\tan y}{\cot x-\cot y}=\dfrac{\tan x-\tan y}{\dfrac{1}{\tan x}-\dfrac{1}{\tan y}}=$

$\displaystyle =\dfrac{\tan x-\tan y}{\dfrac{\tan y-\tan x}{\tan x\tan y}}=\left(\tan x-\tan y\right)\cdot\dfrac{\tan x\tan y}{-\left(\tan x-\tan y\right)}=-\tan x\tan y$
• Feb 2nd 2011, 02:53 AM
ericpepin
Hey where is the R.H.S. and what you trying to prove?
• Feb 3rd 2011, 08:54 AM
Quote:

Originally Posted by youngb11
LS=(tanx-tany)/(cotx-coty)

LS=-tanxtany

If possible try to avoid use of the tan compound formula.

Thanks!

$\displaystyle \displaystyle\frac{tanx-tany}{cotx-coty}=\left(\frac{tanx}{tanx}\right)\frac{tanx-tany}{cotx-coty}=\frac{tanx(tanx-tany)}{1-tanxcoty}$

$\displaystyle =\displaystyle\left(\frac{tany}{tany}\right)\frac{ tanx(tanx-tany)}{1-tanxcoty}=\frac{tanxtany(tanx-tany)}{tany-tanx}$

$\displaystyle =\displaystyle\ -\frac{tanxtany(tany-tanx)}{tany-tanx}$