Prove that (cosxcotx)/(1 - sinx) - 1 = cscx?

2 Answers
May 12, 2016

We start with:

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

First, recall cotx = cosx/sinx (since cotx = 1/tanx and tanx = sinx/cosx). That gives you:

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

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

On the fraction you can move the middle sinx into the denominator.

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

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 = (sinx(1-sinx))/(sinx(1-sinx)) to get:

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 sin^2x + cos^2x = 1 to cancel out the 1-sinx.

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

1/sinx = cscx

color(blue)(cscx = cscx)

May 12, 2016

It is a bit long but...

Explanation:

You can try changing all in sin and cos as:
cot(x)=cos(x)/sin(x)
csc(x)=1/sin(x)
Your identity becomes:

(cos(x)*cos(x)/sin(x))/(1-sin(x))-1=1/sin(x)
we use (1-sin(x))(sin(x)) as common denominator and write rearranging:
(cos^2(x)-sin(x)(1-sin(x)))/cancel((1-sin(x))(sin(x)))=(1-sin(x))/cancel((1-sin(x))(sin(x)))

and:
cos^2(x)-sin(x)(1-sin(x))=1-sin(x)
cos^2(x)-sin(x)+sin^2(x)=1-sin(x)
but:
cos^2(x)+sin^2(x)=1
so we get:
1-sin(x)=1-sin(x) which is true.