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

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

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