# Thread: Help Verifying an Identity

1. ## Help Verifying an Identity

Hi all!

I got myself stuck here trying to figure out how to verify this Identity.

tan x + cot x = sec x csc x

Here is what I have so far

tan x + cot x

1/cot x + cot x/1

(1+cot^2)/cot x

(csc x/cot x) * csc x

(1/sin x)/(1/tan x) * csc x

(tan x/sin x) * csc x

But I can't seem to turn (tan x/sin x) into sec x
thankx for helping

ps i'm still new to this so please be as detailed with your responses as you can thanks.

2. ## Re: Help Verifying an Identity

tan x + cot x = (sin^{2}x + cos^{2}x)/(sin x cos x) = 1 /(sin x cos x) = sec x csc x

3. ## Re: Help Verifying an Identity

What steps did you take to get from

tan x + cot x

to

(sin^{2}x + cos^{2}x)/(sin x cos x) ?

4. ## Re: Help Verifying an Identity

tan x + cot x = tan x * (sin x / sin x) + cot x * (cos x / cos x) = (sin x / cos x) * (sin x / sin x) + (cos x / sin x)* (cos x / cos x)

5. ## Re: Help Verifying an Identity

thank you very much i was able to follow along and see how it was solved.. very helpful

6. ## Re: Help Verifying an Identity

I have another identity I am having trouble with if any1 can help..

(sin x/1-cot x) - (cos x/tan x-1) = sin x + cos x

I was able to clear the denominators but then I am not sure how to go about making the sign positive.

[(sin x/1-cot x)*(tan x-1/tan x-1)] - [(cos x/tan x-1)*(1-cot x/1-cot x)]

= sin x - cos x

thanx

7. ## Re: Help Verifying an Identity

Hello, mathgex!

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

We have: .$\displaystyle \frac{\sin x}{1 - \frac{\cos x}{\sin x}} - \frac{\cos x}{\frac{\sin x}{\cos x} - 1}$

Multiply the first fraction by $\displaystyle \frac{\sin x}{\sin x}$, the second fraction by $\displaystyle \frac{\cos x}{\cos x}$

. . $\displaystyle \frac{\sin x}{\sin x}\cdot\frac{\sin x}{1 - \frac{\cos x}{\sin x}} \:-\: \frac{\cos x}{\cos x}\cdot\frac{\cos x}{\frac{\sin x}{\cos x} - 1}$

. . $\displaystyle =\;\frac{\sin^2\!x}{\sin x - \cos x} - \frac{\cos^2\!x}{\sin x-\cos x} \;=\; \frac{\sin^2\!x - \cos^2\!x}{\sin x - \cos x}$

. . $\displaystyle =\;\frac{(\sin x - \cos x)(\sin x + \cos x)}{\sin x - \cos x} \;=\;\sin x + \cos x$

8. ## Re: Help Verifying an Identity

thanks soroban,

it took me a second to realize the difference of two squares move you made
thank you for helping, whats a good step by step method for solving these things?