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

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Answer 1 · 2015-12-20T06:44:22

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 · 2017-02-01T03:54:47

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.

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