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

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10
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)#

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Gió Share
May 12, 2016

Answer:

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.

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