1. ## Verifying help

Hey everyone. I have a worksheet that I am having some problems with. I have spent the last hour working on these three problems and I have come no where close to the answer. I was absent on the day this was taught so I am a little behind everyone else in my class. ( I tried using Latex but it was not working for me)

sin^3 x-sin^2 x-sin x+1=(-cos^3 x)(sin x-1)

cos x/ 1-sin x = sec x + tan x

(sin x)(cos x)/cos^2 x - sin ^2 x = cot x/ cot^2 x -1

All help is greatly appreciated!

2. ## Re: Verifying help

[TEX]\sin^3 x-\sin^2 x-\sin x+1=(-\cos^3 x)(\sin x-1)[/TEX]

[TEX]\frac{\cos x}{1-\sin x} = \sec x + \tan x[/TEX]

[TEX]\frac{\sin x)(\cos x)}{\cos^2 x - \sin ^2 x} =\frac{\cot x}{\cot^2 x -1}[/TEX]

Gives $\sin^3 x-\sin^2 x-\sin x+1=(-\cos^3 x)(\sin x-1)$

$\frac{\cos x}{1-\sin x} = \sec x + \tan x$

$\frac{\sin x)(\cos x)}{\cos^2 x - \sin ^2 x} =\frac{ cot x}{\cot^2 x -1}$

3. ## Re: Verifying help

TAMH, are you sure for the first equation it isn't supposed to equal $(-\cos^2 x)(\sin x-1)$.

Here are some hints to doing these:

Eq 1) Factor the left by grouping then apply some identities
Eq 2) Multiple the left equation, top and bottom, by the conjugate of the denominator then simplify
Eq 3) Divide the left equation, top and bottom, by $\sin x$ and watch what happens!

4. ## Re: Verifying help

Looking again, I think your third equation may have been written incorrectly too. You sure it's not $\frac{(\sin^2 x)(\cos^2 x)}{\cos^2 x - \sin ^2 x} =\frac{ cos^2 x}{\cot^2 x -1}$

In which case you would divide the top and bottom of the expression on the left by $sin^2 x$ and THEN see what happens.