Question

Using `sin^2x+cos^2x=1` and other trig properties, simplify the following: `{cos^2 x tan^2 x }/{ 1-sin^2 x}` ? Show all work.

Answer 1

`{cos^2 x tan ^2 x}/{1 - sin ^2 x} ={ \cos ^2 x \ {sin ^2 x}/{cos ^2 x} }/{ cos ^2 x} = {sin ^2 x}/{cos^2x } = tan ^2 x`

Explanation:

Show all work? Goes without saying.

Not too much work here. We substitute

`tan^2 x = sin ^2 x / cos ^2 x`

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

`1 - sin ^2 x = cos ^2 x`

So

`{cos^2 x tan ^2 x}/{1 - sin ^2 x} ={ \cos ^2 x \ {sin ^2 x}/{cos ^2 x} }/{ cos ^2 x} = {sin ^2 x}/{cos^2x } = tan ^2 x`

Answer 2

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

Explanation:

We have,

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

`=>cos^2x=1-sin^2x`

So,

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