# How do you prove the identity (1-sinx)/cosx=cosx/(1+sinx)?

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#### Explanation

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#### Explanation:

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16
sjc Share
Oct 15, 2016

By multiplying both numerator and denominator by $1 + \sin x$ and using the difference of squares the result follows quickly.

#### Explanation:

multiply the LHS , top and bottom by $\left(1 + \sin x\right)$

$\frac{\left(1 - \sin x\right) \left(1 + \sin x\right)}{\cos x \left(1 + \sin x\right)}$

$= \frac{1 - {\sin}^{2} x}{\cos x \left(1 + \sin x\right)}$

but ${\sin}^{2} x + {\cos}^{x} = 1$

$\therefore = \frac{{\cos}^{2} x}{\cos x \left(1 + \sin x\right)}$

$= \frac{\cancel{\cos x} \left(\cos x\right)}{\cancel{\cos} x \left(1 + \sin x\right)}$

as required.

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

5
Oct 16, 2016

See the Explanation.

#### Explanation:

We have, $1 - {\sin}^{2} x = {\cos}^{2} x \Rightarrow \left(1 - \sin x\right) \left(1 + \sin x\right) = \left(\cos x\right) \left(\cos x\right)$

$\Rightarrow \frac{1 - \sin x}{\cos} x = \cos \frac{x}{1 + \sin x}$.

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