Question

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

Answer 1

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.

Answer 2

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.