How do you show that #(cscx - cotx)^2 = (1 - cosx)/(1 + cosx)#?

2 Answers
Apr 11, 2017

Convert to sine and cosine using #cscx = 1/sinx# and #cotx= cosx/sinx#.

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

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

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

Now use #sin^2x+ cos^2x = 1#.

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

Note the difference of squares in #1 - cos^2x#, such that #1 - cos^2x = (1 + cosx)(1- cosx)#.

#(1- cosx)^2/((1 + cosx)(1 - cosx)) = (1 - cosx)/(1 + cosx)#

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

This identity has been proven to be true.

Hopefully this helps!

Apr 11, 2017

Multiply the RHS by #(1-cosx)/(1-cosx)#; distribute, and use Pythagorean and reciprocal identities.

Explanation:

Whenever you see a trig identity proof that has something like #(1 +- "trig " x)# in a denominator (where "trig" is one of the 6 trig functions), a good first thing to try is multiplying it by its conjugate.

In our case, #(1+cosx)# is in a denominator, so we choose to multiply it by its conjugate, #(1-cosx)#. (To do this, of course, we need to multiply the numerator by it as well, so we're essentially multiplying the fraction by a fancy version of #1#, which does not change its value.)

#(1-cosx)/(1+cosx)=(1-cosx)/(1+cosx) color(red)(xx (1-cosx)/(1-cosx))#

#color(white)((1-cosx)/(1+cosx))=(1-cosx)^2/(1-cos^2x)"             "#(distribution)

#color(white)((1-cosx)/(1+cosx))=(1-cosx)^2/(sin^2x)"             "#(Pythagorean identity)

#color(white)((1-cosx)/(1+cosx))=((1-cosx)/sinx)^2"        "["thing"^2/"stuff"^2=("thing"/"stuff")^2]#

#color(white)((1-cosx)/(1+cosx))=(1/sinx-cosx/sinx)^2"     "#(splitting the fraction)

#color(white)((1-cosx)/(1+cosx))=(cscx-cotx)^2"           "#(reciprocal identities)

Note:

You can also convert everything to #sin#'s and #cos#'s. This is sort of a "catch-all", like how the quadratic formula can be used to solve any quadratic equation (of one variable). It will always work, but there might be a faster way.