Question

Can anyone help me with these two problems?

Answer 1

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$\dfrac{\sin ^2 x + \sin x \cos x}{\cos x - 2 \cos^3 x} = \dfrac{\tan x}{\sin x - \cos x}$

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.

$\dfrac{\sin ^2 x + \sin x \cos x}{\cos x - 2 \cos^3 x}$

$=\dfrac{\sin x}{\cos x} \cdot \dfrac{ \sin x + \cos x}{1 - 2 \cos^2 x}$

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.

$= \tan x \dfrac{ \sin x + \cos x}{1 - \cos^2 x - (1 - \sin ^2 x)}$

$= \tan x \dfrac{ \sin x + \cos x}{\sin^2 x - \cos^2 x}$

$= \tan x \dfrac{ \sin x + \cos x}{(\sin x + \cos x)(\sin x - \cos x)}$

$=\dfrac{\tan x}{\sin x - \cos x} \quad\checkmark$

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