how to simplify (1/sin^2x) - (1/tan^2x)?

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Answer 1 · 2018-04-06T19:51:47

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

#=sin^2x/sin^2x=1#.

Or, #1/sin^2x-1/tan^2x=csc^2x-cot^2x=1#.

Answer 2 · 2018-04-06T19:53:16

It happens to be exactly #1#.

#(1/sin^2x) - (1/tan^2x)#

#=(1/sin^2x) - (1/\frac{sin^2x}{\cos^2x})#

#=1/sin^2x - \frac{\cos^2x}{sin^2x}#

#=\frac{1-cos^2x}{sin^2x}#

#=\frac{sin^2x}{sin^2x} = 1#

Note that this uses the Pythagorean trigonometric identity in the second-to-last step, which states that:

#sin^2x+cos^2x=1#