# Math Help - Trig Indentity Help (Proof)

1. ## Trig Indentity Help (Proof)

$[cot X/(1-tan X)] + [tan X/(1-cot X] = 1+ tan X + cot X$

2. I tried this with no luck... tried multiplying out by the conjugates and then LCD's, but nothing...

3. Originally Posted by viciouspoultry
$[cot X/(1-tan X)] + [tan X/(1-cot X] = 1+ tan X + cot X$
$\frac{{\cot x}}
{{1 - \tan x}} + \frac{{\tan x}}
{{1 - \cot x}} = \frac{{\dfrac{{\cos x}}
{{\sin x}}}}
{{\dfrac{{\cos x - \sin x}}
{{\cos x}}}} - \frac{{\dfrac{{\sin x}}
{{\cos x}}}}
{{\dfrac{{\cos x - \sin x}}
{{\sin x}}}}.$

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

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

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

$1 + \tan x + \cot x,$ as required $\blacksquare$

4. Hello, viciouspoultry!

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

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

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

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

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

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

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