Question

Simplify the expression.

(csc^2xsec^2x)/(sec^2x+csc^2x)

Answer 1

do we put everyting in terms of sin and cos?

Answer 2

Yup. do that and you would be able to simplify it.

Answer 3

I get to that point and IDK what to do though

Answer 4

I usually do. You don't have to but I find that easiest.

Answer 5

What do you have?

Answer 6

(1/sin^2x)(1/cos^2x) all divided by 1/cos^2x + 1/sin^2x

Answer 7

Now leave the numertor as it is. Simplify the denominator first.

Answer 8

Ok now combine the two fractions in the denominator

Answer 9

1/cos^2x+sin^2x=1/1? os 1 on the denominator?

Answer 10

You'll take the LCM right?

so you'll get sin^2 x + cos^2 x/(sin^2*cos^2 x)..

Answer 11

no im confused...

Answer 12

well the denominator was ...\[\frac{cos^2x+sin^2x}{sin^2xcos^2}\]

Answer 13

how?

Answer 14

how would you solve this..
1/2 + 1/3?

Answer 15

LCM but im confused i though the denominator was 1/cos^2x+sin^2x

Answer 16

you're getting the second term wrong..cosec^2 x= 1/sin^2 x

Answer 17

thats what I have

Answer 18

ok i understand...you have to multiply each fraction by the other fractions cos or sin

Answer 19

so they get the samedenominator

Answer 20

yes

Answer 21

so then for the denominator you have 1/sin^2cos^2

Answer 22

is the answer 1?

Answer 23

Exactly. so you get sec^4 x* cosec^4x
Yeah? I skipped a step.

Answer 24

you have to rationalize the denominator, so you multiply by sin^2xcos^2x on top and bottom and then everything multiplies and cancels out on top?

Answer 25

Oh dang. yeah i forgot we had sin and cos in the denominator. Well done.

Answer 26

Yes that is correct the final answer is 1! Nice job!

Answer 27

So im right??? :D YAY! Thanks!

Answer 28

1 is correct: http://www.wolframalpha.com/input/?i=%28csc%5E2xsec%5E2x%29%2F%28sec%5E2x%2Bcsc%5E2x%29