Help with verifying trig functions please!

**TacticalPro** · February 8th 2011, 05:21 PM

Hi, I'm having some trouble verifying a trig equation using identities. here is the problem:

show that:

(tanx-cotx)/(sinxcosx) = sec^2(x) - csc^2(x)

i'm stuck, but I think the first thing I have to do is the pythagorean identity with tanx and cotx. is this right? if so what do i do next? (I'm working on left side of equation)

**dwsmith** · February 8th 2011, 05:41 PM

Originally Posted by TacticalPro

Hi, I'm having some trouble verifying a trig equation using identities. here is the problem:

show that:

(tanx-cotx)/(sinxcosx) = sec^2(x) - csc^2(x)

i'm stuck, but I think the first thing I have to do is the pythagorean identity with tanx and cotx. is this right? if so what do i do next? (I'm working on left side of equation)

$\displaystyle\frac{\tan x-\cot x}{\sin x\cos x}=\frac{\tan x}{\sin x\cos x}-\frac{\cot x}{\sin x\cos x}=\frac{\frac{\sin x}{\cos x}}{\sin x\cos x}\cdots$

Get the idea?

**Ithaka** · February 8th 2011, 05:46 PM

Originally Posted by TacticalPro

Hi, I'm having some trouble verifying a trig equation using identities. here is the problem:

show that:

(tanx-cotx)/(sinxcosx) = sec^2(x) - csc^2(x)

i'm stuck, but I think the first thing I have to do is the pythagorean identity with tanx and cotx. is this right? if so what do i do next? (I'm working on left side of equation)

$\frac{\tan x-\cot x}{\sin x * \cos x}$

$=\frac{\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}}{\sin x * \cos x}$

$=\frac{\frac{\sin ^2 x - \cos ^2 x}{\sin x * \cos x}}{\sin x * \cos x}$

$=\frac{\sin ^2 x - \cos ^2 x}{\sin ^2 x * \cos^2 x}$

$=\frac{\sin ^2 x}{\sin ^2 x * \cos^2 x}-\frac{\cos ^2 x}{\sin ^2 x * \cos^2 x}$

Simplify it and you will get the right hand side of your identity.

**TacticalPro** · February 8th 2011, 06:02 PM

Originally Posted by Ithaka

$\frac{\tan x-\cot x}{\sin x * \cos x}$

$=\frac{\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}}{\sin x * \cos x}$

$=\frac{\frac{\sin ^2 x - \cos ^2 x}{\sin x * \cos x}}{\sin x * \cos x}$

$=\frac{\sin ^2 x - \cos ^2 x}{\sin ^2 x * \cos^2 x}$

$=\frac{\sin ^2 x}{\sin ^2 x * \cos^2 x}-\frac{\cos ^2 x}{\sin ^2 x * \cos^2 x}$

Simplify it and you will get the right hand side of your identity.

Thanks!

I originally did that but got stuck at the end because i forgot I could split the fraction into two parts

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