1. ## difficult identity

I've been working on this problem for over an hour, and I still am unable to verify it. I know it is true, because I plugged in a value for x and it works, but i don't know how to prove it.

(tan x / (1 - cot x)) + (cot x / (1-tan x))^2 = 1 + sec x csc x

Thanks

2. Originally Posted by qwerty723519@hotmail.com
I've been working on this problem for over an hour, and I still am unable to verify it. I know it is true, because I plugged in a value for x and it works, but i don't know how to prove it.

(tan x / (1 - cot x)) + (cot x / (1-tan x))^2 = 1 + sec x csc x

Thanks
start with the left hand side. combine the fractions, change everything to sines and cosines and simplify as much as possible.

then go to the left hand side, change everything to sines and cosines and show that you can simplify it to get the same thing

3. yea, I know thats the method of solving them, and I was actually able to solve 19 out of the 20 I had to do, but this one just doesn't seem to simplify.

4. Originally Posted by qwerty723519@hotmail.com
yea, I know thats the method of solving them, and I was actually able to solve 19 out of the 20 I had to do, but this one just doesn't seem to simplify.
just to clarify, is the problem $\frac {\tan x}{1 - \cot x} + \left( \frac {\cot x}{1 - tan x} \right)^2 = 1 + \sec x \csc x$ ?

5. ⁡No it is actually:

〖tan⁡x/(1-cot⁡x )〗+ cot⁡x/(1-tan⁡x )=1+ sec⁡〖x csc⁡x 〗

ps. I typed the formula into microsoft word using the symbols, so that it was easier to visualize, but it didn't copy over the same. How did you write that formula?

6. Originally Posted by qwerty723519@hotmail.com
⁡No it is actually:

〖tan⁡x/(1-cot⁡x )〗+ cot⁡x/(1-tan⁡x )=1+ sec⁡〖x csc⁡x 〗

ps. I typed the formula into microsoft word using the symbols, so that it was easier to visualize, but it didn't copy over the same. How did you write that formula?
hehe, what you write now is even more confusing than the last one. what do the ?'s mean?

i used LaTeX. see the LaTeX tutorial here to see how

7. whoops, sorry. It doesn't display ?s for me, but I'll try this LaTex thing, which might take me a little while.

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

whew, that was rough.

8. Hello, qwerty!

$\frac {\tan x}{1 - \cot x} + \frac {\cot x}{1 - \tan x} \:= \:1 + \sec x \csc x$

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

Multiply each fraction by $\frac{\sin x\cos x}{\sin x\cos x}$

. . $\frac{\sin x\cos x}{\sin x\cos x}\cdot\frac{\dfrac{\sin x}{\cos x}}{1 - \dfrac{\cos x}{\sin x}} \;+ \;\frac{\sin x\cos x}{\sin x\cos x}\cdot\frac{\dfrac{\cos x}{\sin x}}{1 - \dfrac{\sin x}{\cos x}} \;=$ . $\frac{\sin^2\!x}{\sin x\cos x - \cos^2\!x} + \frac{\cos^2\!x}{\sin x\cos x - \sin^2\!x}$

. . $= \;\frac{\sin^2\!x}{\cos x(\sin x-\cos x)} - \frac{\cos^2\!x}{\sin x(\sin x - \cos x)} \;=\;\frac{\sin^3\!x - \cos^3\!x}{\cos x\sin x(\sin x-\cos x)}$

Factor: . $\frac{(\sin x-\cos x)(\sin^2\!x + \sin x\cos x + \cos^2\!x)}{\cos x\sin x(\sin x-\cos x)} \;=\;\frac{\sin x\cos x + \overbrace{\sin^2\!x + \cos^2\!x}^{\text{This is 1}}}{\cos x\sin x}$

. . $= \;\frac{\sin x\cos x + 1}{\cos x\sin x} \;=\;\frac{\sin x\cos x}{\cos x\sin x} + \frac{1}{\cos x\sin x} \;=\;1 + \sec x\csc x$

9. wow thank you so much! I would've spent another 2 hours on that to no avail, because I didn't even think of using difference of cubes.