Help Solving Precalculus Cot x + Tan x + 1 Problem

[pasatom20]

Can anyone help me sove this problem?

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

 

[courtrigrad]

convert the right hand sides to sines and cosines, add the fractions, and simplify. or you could use the fact that tan x = 1 / cot x

 

[lkh1986]

Since right hand side is more complex, I think we should start from that side, then arrive at the left hand side.

We can use the properties of that cot x is the reciprocal of tan x, too.

 

[pasatom20]

Here is what i got by working on the right hand side.

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

then i tried several different ways to work on the problem from this step, but it never worked.

 

[courtrigrad]

convert the numerators to sines and cosines as well.

 

[pasatom20]

here is what i got

(cos x / sin x ) / (cos x - sin x / cos x ) + (sin x / cos x) / (sin x - cos x / sin x)
= (cos x / sin x )*(cos x / cos x - sin x) + (sin x / cos x)*(sin x / sin x - cos x)
= (cos ^2 x / cos x - sin^2 x) + (sin^2 x / sin x - cos^2 x)
= (sin^2 x - 1 / cos x - sin^2 x ) + (cos^2 x - 1 / sin x - cos^2 x)
I tried to work on the problem until i got to this step, but I'm not sure if I'm on the right track.

 

[courtrigrad]

use the identity tan x = 1 / cot x instead.

So \cot x + \tan x + 1 = \frac{\cot x}{1- \frac{1}{\cot x}} etc..