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

Question

Ariana Alvarez · Accepted Answer

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)#

Anna Allen · Answer

It is a bit long but...

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.

Elias Ackerman · Answer

undefined To prove the identity:

(cos(x) * cot(x)) / (1 - sin(x)) - 1 = csc(x)

We start with the left side and manipulate it to match the right side.

(cos(x) * cot(x)) / (1 - sin(x)) - 1

= (cos(x) * (cos(x) / sin(x))) / (1 - sin(x)) - 1

= (cos^2(x) / sin(x)) / (1 - sin(x)) - 1

= (cos^2(x) / sin(x)) / ((1 - sin(x)) / 1) - 1

= cos^2(x) / sin(x) * (1 / (1 - sin(x))) - 1

= cos^2(x) / sin(x) * csc(x) - 1

= (cos^2(x) / sin(x)) * (1 / sin(x)) - 1

= (cos^2(x) * (1 / sin(x)^2)) - 1

= (cos^2(x) * csc^2(x)) - 1

Now, using the Pythagorean identity for cosine:

cos^2(x) = 1 - sin^2(x)

= ((1 - sin^2(x)) * csc^2(x)) - 1

= (csc^2(x) - sin^2(x) * csc^2(x)) - 1

= (csc^2(x) - 1) - 1

= csc^2(x) - 2

= csc^2(x) - 1 + (-1)

= csc^2(x) - 1

= csc(x)

Therefore, we've proven that (cos(x) * cot(x)) / (1 - sin(x)) - 1 = csc(x).