How do you simplify #(csc² t - cot² t) / sin² t#?

1 Answer
Jake M.
Mar 4, 2018

#sec^2 t#

Explanation:

I like just sines and cosines, so let's work with those:

#(csc^2 t - cot^2 t)/(sin^2(t))= (1/(sin^2t) - (cos^2t)/(sin^2t))/(sin^2t) #

Let's get rid of the fractions up top by multiplying top and bottom by #sin^2t#

#= (1 - cos^2t)/(sin^4 t)#

Because #sin^2t + cos^2 t = 1#, #1- cos^2 t = sin^2 t#.
#= (sin^2t)/(sin^4t) = 1/(sin^2t) = sec^2 t #

In hindsight, we could have also realized that there is a trig identity that
#csc^2t-cot^2t = 1 #
but that involves memorization, whereas the first thing doesn't.

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