How do you simplify #(sin² t - cos² t sin² t) / sin² t#?

1 Answer
Nov 5, 2015

#sin^2t#

Explanation:

First off, I am lead to believe the "t" outside the denominator is a mistake. Secondly, for my convenience I changed "t" to "x".

1) Since in this equation the numerator has a subtraction we could and will split the numerator.
2) Now that the numerator is split we can simplify #sin^2t/sin^2t# into 1. As well as cancel #sin^2t# in both the numerator and denominator for the other fraction.
3) This leaves us with #1-cos^2t#, which if you notice is the Pythagorean Identity of #sin^2x+cos^2x=1#, but is can be rewritten as #sin^2x=1-cos^2x#. So now that we know it is an identity we can simplify it as #sin^2t#.
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