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[tex]\dfrac{\sin ^2 x + \sin x \cos x}{\cos x - 2 \cos^3 x} = \dfrac{\tan x}{\sin x - \cos x}[/tex]
I don't like this computer aided instruction holding my hand. I'll ignore that at first so I can compare my solution to the one being taught.
We can see how a tan factors out; let's see what's left.
[tex]\dfrac{\sin ^2 x + \sin x \cos x}{\cos x - 2 \cos^3 x}[/tex]
[tex]=\dfrac{\sin x}{\cos x} \cdot \dfrac{ \sin x + \cos x}{1 - 2 \cos^2 x}[/tex]
We need to factor out a sin x + cos x in the denominator so let's change one of the squared cosines to squared sine.
[tex]= \tan x \dfrac{ \sin x + \cos x}{1 - \cos^2 x - (1 - \sin ^2 x)}[/tex]
[tex]= \tan x \dfrac{ \sin x + \cos x}{\sin^2 x - \cos^2 x}[/tex]
[tex]= \tan x \dfrac{ \sin x + \cos x}{(\sin x + \cos x)(\sin x - \cos x)}[/tex]
[tex]=\dfrac{\tan x}{\sin x - \cos x} \quad\checkmark[/tex]
OK, now I can look at what they did.
First step ok, the box is 2 cos^2 x
Second step they changed the 1, box same as before: 2 cos^2 x
Third box matches our proof, cos^2 x
Fourth box our last box didn't change: cos^2 x
Fifth box: sin x + cos x
