# Trig Indentity Help (Proof)

• Mar 5th 2008, 09:56 AM
viciouspoultry
Trig Indentity Help (Proof)
$\displaystyle [cot X/(1-tan X)] + [tan X/(1-cot X] = 1+ tan X + cot X$
• Mar 5th 2008, 10:52 AM
colby2152
I tried this with no luck... tried multiplying out by the conjugates and then LCD's, but nothing...
• Mar 5th 2008, 11:46 AM
Krizalid
Quote:

Originally Posted by viciouspoultry
$\displaystyle [cot X/(1-tan X)] + [tan X/(1-cot X] = 1+ tan X + cot X$

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

Now this is equal to $\displaystyle \frac{{\cos ^2 x}} {{\sin x(\cos x - \sin x)}} - \frac{{\sin ^2 x}} {{\cos x(\cos x - \sin x)}}.$ Factorise:

$\displaystyle \frac{1} {{\cos x - \sin x}}\left\{ {\frac{{\cos ^2 x}} {{\sin x}} - \frac{{\sin ^2 x}} {{\cos x}}} \right\}.$ After some simple calculations we have

$\displaystyle \frac{1} {{\cos x - \sin x}}\left\{ {\frac{{(\cos x - \sin x)\left( {\cos ^2 x + \sin x\cos x + \sin ^2 x} \right)}} {{\sin x\cos x}}} \right\},$ and we happily get

$\displaystyle 1 + \tan x + \cot x,$ as required $\displaystyle \blacksquare$
• Mar 5th 2008, 03:21 PM
Soroban
Hello, viciouspoultry!

Thought I'd try this one head-on . . .

Quote:

$\displaystyle \frac{\cot x}{1-\tan x}+\frac{\tan x}{1-\cot x} \;= \;1+ \tan x + \cot x$

On the left, multiply the second fraction by $\displaystyle \frac{\tan x}{\tan x}\!:\;\;\;\frac{\cot x}{1 -\tan x} \:+\:{\color{blue}\frac{\tan x}{\tan x}}\cdot\frac{\tan x}{1-\cot x}$

. . $\displaystyle =\;\;\frac{\cot x}{1-\tan x} + \frac{\tan^2\!x}{\tan x - 1}\;\;=\;\;\frac{\cot x}{1-\tan x} - \frac{\tan^2\!x}{1-\tan x} \;\;=\;\; \frac{\cot x-\tan^2\!x}{1-\tan x}$

Multiply by $\displaystyle \frac{\tan x}{\tan x}\!:\;\;\;{\color{blue}\frac{\tan x}{\tan x}}\cdot\frac{\cot x - \tan^2\!x}{1-\tan x} \;=\; \frac{1 - \tan^3\!x}{\tan x(1-\tan x)}$

Factor and reduce: .$\displaystyle \frac{(1-\tan x)(1+ \tan x+ \tan^2\!x)}{\tan x(1-\tan x)} \;=\;\frac{1+\tan x+\tan^2\!x}{\tan x}$

Then we have: . $\displaystyle \frac{1}{\tan x} + \frac{\tan x}{\tan x} + \frac{\tan^2\!x}{\tan x}\;\;=\;\; \cot x + 1 +\tan x$