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

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#### Explanation

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We start with:

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

First, recall

#(cosx*cosx/sinx)/(1-sinx) - 1 = cscx#

#(cos^2x/sinx)/(1-sinx) - 1 = cscx#

On the fraction you can move the middle

#cos^2x/(sinx(1-sinx)) - 1 = cscx#

Now, when you have a

#cos^2x/(sinx(1-sinx)) - (sinx(1-sinx))/(sinx(1-sinx)) = cscx#

Distribute the numerator and combine into one fraction:

#(-sinx + cos^2x + sin^2x)/(sinx(1-sinx)) = cscx#

Then recall

#cancel(1 - sinx)/(sinxcancel((1-sinx))) = cscx#

#1/sinx = cscx#

#color(blue)(cscx = cscx)#

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Describe your changes (optional) 200

#### Answer:

It is a bit long but...

#### Explanation:

You can try changing all in

Your identity becomes:

we use

and:

but:

so we get:

Describe your changes (optional) 200