How do you verify #cscx-sinx=cosxcotx#?

1 Answer
Apr 16, 2016

Let's start by stating the identities that will be important to this problem:

Explanation:

The reciprocal identity (1): #cscx = 1/sinx#

The quotient identity (1): #cotx = cosx/sinx#

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

Placing the left side on a common denominator:

#1/sinx - sin^2x/sinx = cos^2x/sinx#

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

Applying the Pythagorean identity #cos^2x + sin^2x = 1#, we get:

#cos^2x/sinx = cos^2x /sinx#

Hopefully this helps!