How do you simplify # (1- sin^2 theta)/( csc^2 theta -1)#?

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Answer 1 · 2016-08-02T09:28:52

#sin^2theta#

Except when #theta = pi/2 + npi, n in ZZ# (See Zor's explanation)

Lets look at the numerator and denominator separately first.

#1-sin^2theta = cos^2theta#

#csc^2theta = 1/(sin^2theta)#

#1/(sin^2theta) - 1 = (1-sin^2theta)/(sin^2theta) = (cos^2theta)/(sin^2theta)#

So

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