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Iteru [2.4K]
3 years ago
13

How to convert it into seconds?

Mathematics
1 answer:
lys-0071 [83]3 years ago
4 0

Answer:

What are we supposed to covert?

Step-by-step explanation:

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Can anyone help me with these two problems?
Veseljchak [2.6K]

Show

\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.

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6 0
4 years ago
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