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!

How do you simplify the expression 1/sin^2A-1/tan^2A1/sin^2A-1/tan^2A?