Solving a trig identity

**youngb11** · November 20th 2010, 08:32 PM

LS=(tanx-tany)/(cotx-coty)

LS=-tanxtany

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

Thanks!

**dwsmith** · November 20th 2010, 08:33 PM

What are you trying to do?

**red_dog** · November 21st 2010, 12:08 AM

$\dfrac{\tan x-\tan y}{\cot x-\cot y}=\dfrac{\tan x-\tan y}{\dfrac{1}{\tan x}-\dfrac{1}{\tan y}}=$

$=\dfrac{\tan x-\tan y}{\dfrac{\tan y-\tan x}{\tan x\tan y}}=\left(\tan x-\tan y\right)\cdot\dfrac{\tan x\tan y}{-\left(\tan x-\tan y\right)}=-\tan x\tan y$

**ericpepin** · February 2nd 2011, 02:53 AM

Hey where is the R.H.S. and what you trying to prove?

**Archie Meade** · February 3rd 2011, 08:54 AM

Originally Posted by youngb11

LS=(tanx-tany)/(cotx-coty)

LS=-tanxtany

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

Thanks!

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

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

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

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