3 years ago

What angle relationship describes angles CBE and DEB?

Mathematics

1 answer:

Archy [21]3 years ago

3 0

Answer:

This is what it shows

Step-by-step explanation:

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Answer:

See below

Step-by-step explanation:

We want to prove that

$\sin(x)\tan(x) = \dfrac{1}{\cos(x)} - \cos(x), \forall x \in\mathbb{R}$

Taking the RHS, note

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

Remember that

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

Once

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