# How do you verify trigonometric identities: (cosx*cotx)/(1-sinx)-1=cscx ?

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10
May 12, 2016

color(green)((cosxcotx)/(1-sinx) - 1 stackrel(?)(=) cscx).

First, recall $\cot x = \cos \frac{x}{\sin} x$ (since $\cot x = \frac{1}{\tan} x$ and $\tan x = \sin \frac{x}{\cos} x$). That gives you:

$\frac{\cos x \cdot \cos \frac{x}{\sin} x}{1 - \sin x} - 1 = \csc x$

$\frac{{\cos}^{2} \frac{x}{\sin} x}{1 - \sin x} - 1 = \csc x$

On the fraction you can move the middle $\sin x$ into the denominator.

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

Now, when you have a $1$, that gives you a lot of freedom. You can choose to have it be equal to any ratio you want, as long as it cancels out to be $1$. So choose $1 = \frac{\sin x \left(1 - \sin x\right)}{\sin x \left(1 - \sin x\right)}$ to get:

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

Distribute the numerator and combine into one fraction:

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

Then recall ${\sin}^{2} x + {\cos}^{2} x = 1$ to cancel out the $1 - \sin x$.

$\frac{\cancel{1 - \sin x}}{\sin x \cancel{\left(1 - \sin x\right)}} = \csc x$

$\frac{1}{\sin} x = \csc x$

$\textcolor{b l u e}{\csc x = \csc x}$

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

2
Gió Share
May 12, 2016

It is a bit long but...

#### Explanation:

You can try changing all in $\sin$ and $\cos$ as:
$\cot \left(x\right) = \cos \frac{x}{\sin} \left(x\right)$
$\csc \left(x\right) = \frac{1}{\sin} \left(x\right)$

$\frac{\cos \left(x\right) \cdot \cos \frac{x}{\sin} \left(x\right)}{1 - \sin \left(x\right)} - 1 = \frac{1}{\sin} \left(x\right)$
we use $\left(1 - \sin \left(x\right)\right) \left(\sin \left(x\right)\right)$ as common denominator and write rearranging:
$\frac{{\cos}^{2} \left(x\right) - \sin \left(x\right) \left(1 - \sin \left(x\right)\right)}{\cancel{\left(1 - \sin \left(x\right)\right) \left(\sin \left(x\right)\right)}} = \frac{1 - \sin \left(x\right)}{\cancel{\left(1 - \sin \left(x\right)\right) \left(\sin \left(x\right)\right)}}$

and:
${\cos}^{2} \left(x\right) - \sin \left(x\right) \left(1 - \sin \left(x\right)\right) = 1 - \sin \left(x\right)$
${\cos}^{2} \left(x\right) - \sin \left(x\right) + {\sin}^{2} \left(x\right) = 1 - \sin \left(x\right)$
but:
${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$
so we get:
$1 - \sin \left(x\right) = 1 - \sin \left(x\right)$ which is true.

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