# I having a headache proving an identity

• Mar 30th 2009, 08:28 AM
darkmoon123
I having a headache proving an identity
I having some problem proving a identity.

Prove:
$\displaystyle (sec^2x )/tanx = (cot^2x - tan^2x) / (cotx - tanx)$

i start by the left side so i get (1+tan^2x) / tan
, and i then i convert tans to sins and cons, so i get (1+(sinX^2/con^2x))/(sinx/sin)

after that i am lost, i dont know if i should start by the left side.

plz help
• Mar 30th 2009, 08:42 AM
TheEmptySet
Quote:

Originally Posted by darkmoon123
I having some problem proving a identity.

Prove:
$\displaystyle (sec^2x )/tanx = (cot^2x - tan^2x) / (cotx - tanx)$

i start by the left side so i get (1+tan^2x) / tan
, and i then i convert tans to sins and cons, so i get (1+(sinX^2/con^2x))/(sinx/sin)

after that i am lost, i dont know if i should start by the left side.

plz help

I would start with the left hand side b/c it is the difference of squares

$\displaystyle \frac{(\cot(x)+tan(x))(\cot(x)-\tan(x))}{\cot(x)-tan(x)}=\cot(x)+tan(x)=\frac{\cos(x)}{\sin(x)}+\fr ac{\sin(x)}{\cos(x)}$

$\displaystyle =\frac{\cos^{2}(x)+\sin^2(x)}{\sin(x)\cos(x)}=\fra c{1}{\sin(x)\cos(x)}\cdot \left( \frac{\frac{1}{\cos(x)}}{\frac{1}{\cos(x)}}\right) =\frac{\sec^2(x)}{\tan(x)}$
• Mar 30th 2009, 09:07 AM
Soroban
Hello, darkmoon123!

Quote:

Prove: .$\displaystyle \frac{\sec^2x}{\tan x} \:=\:\frac{\cot^2\!x - \tan^2\!x}{\cot x - \tan x}$

$\displaystyle \frac{\cot^2\!x-\tan^2\!x}{\cot x -\tan x} \;=\;\frac{(\cot x-\tan x)(\cot x + \tan x)}{\cot t - \tan x} \;=\;\cot x + \tan x$

. . $\displaystyle = \;\frac{1}{\tan x} + \tan x \;=\;\frac{\overbrace{1+\tan^2\!x}^{\text{This is }\sec^2\!x}}{\tan x} \;=\;\frac{\sec^2\!x}{\tan x}$

• Mar 30th 2009, 09:44 AM
darkmoon123
http://www.mathhelpforum.com/math-he...5a62f7b3-1.gif

I actually wanted to resolve the hardest part which is is the left compressed part, so i need to expand until it looks like the right part.

$\displaystyle (1+tan^2x) / (tanx)$

$\displaystyle = ((cos^2x ) / (cos^2x) + (sin^2x) / (cos^2x)) / (sinx)/(cos)$

but it fails me.
• Mar 30th 2009, 10:00 AM
e^(i*pi)
Quote:

Originally Posted by darkmoon123
http://www.mathhelpforum.com/math-he...5a62f7b3-1.gif

I actually wanted to resolve the hardest part which is is the left compressed part, so i need to expand until it looks like the right part.

$\displaystyle (1+tan^2x) / (tanx)$

$\displaystyle = ((cos^2x ) / (cos^2x) + (sin^2x) / (cos^2x)) / (sinx)/(cos)$

but it fails me.

$\displaystyle \frac{1+tan^2(x)}{tan(x)} = \frac{1}{tan(x)} + \frac{tan^2(x)}{tan(x)} = cot(x) + tan(x)$

Multiply by $\displaystyle \frac{cot(x)-tan(x)}{cot(x)-tan(x)}$ which will give the same denominator as the right hand side.

$\displaystyle \frac{(cot(x)+tan(x))(cot(x)-tan(x))}{cot(x)-tan(x)}$

noting that the numerator is the difference of two squares we get the RHS

$\displaystyle \frac{(cot^2x - tan^2x)}{(cotx - tanx)}$

I can't take the credit for this, it's essentially what TheEmptySet did in reverse