How do you simplify the expression #1/sin^2A-1/tan^2A#?

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Answer 1 · 2016-08-29T12:43:20

#=1#

#1/sin^2A-1/tan^2A#

#=1/sin^2A-cot^2A#

#=1/sin^2A-cos^2A/sin^2A#

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

#=sin^2A/sin^2A#

#=1#

Answer 2 · 2016-08-29T14:59:27

#1#

Alternate approach.

Start by applying the identity #tanalpha = sin alpha/cosalpha#.

#=>1/(sin^2a) - 1/(sin^2A/cos^2A)#

Now, simplify:

#=>1/(sin^2a) - cos^2A/sin^2A#

#=> (1 - cos^2A)/sin^2A#

Now, rearrange the pythagorean identity #cos^2beta + sin^2beta = 1#, solving for #sin^2beta# to get #sin^2beta = 1- cos^2beta#:

#=> sin^2A/sin^2A#

Cancel using the property #a/a = 1, a !=0#

#=> 1#

Hopefully this helps!