How do you prove #(cosx+ 1)/ sin^3x = cscx/(1-cosx)#?

2 Answers
Dec 20, 2015

Let's manipulate only the right hand side in order to make it appear like the left hand side.

First, multiply by the conjugate of the denominator.

#cscx/(1-cosx)*(1+cosx)/(1+cosx)=(cscx(1+cosx))/(1-cos^2x)#

Note that since #sin^2x+cos^2x=1#, it's true that #sin^2x=1-cos^2x#.

Thus, the expression can be rewritten as:

#=(cscx(cosx+1))/sin^2x#

Recall that #cscx=1/sinx#.

#=(1/sinx(cosx+1))/sin^2x#

Multiply the numerator and denominator by #sinx#.

#=(1/sinx(cosx+1))/sin^2x*sinx/sinx#

#=(cosx+1)/sin^3x#

Voilà! This is the left hand side.

Feb 1, 2017

This is method without conjugates. It involves manipulating only the left-hand side of the equation.

#(cosx+1)/sin^3x#

#=(cosx+1)/(sinx(sin^2x))#

Through the Pythagorean Identity:

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

Factoring:

#=(cosx+1)/(sinx(1-cosx)(1+cosx))#

Cancelling #(cosx+1)/(1+cosx)=(cosx+1)/(cosx+1)=1#:

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

Rewriting #sinx# as #1/cscx#:

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

Inverting:

#=cscx/(1-cosx)#

Which is the right-hand side. Thus the equality is proven.