Question

Simplify the expression:

cos^2 x + sin^2 x
---------------------
cot^2 x + csc^2 x

Okay, I still suck at Identities here, so if you could tell me if I'm on the right track, that'd be great :)

So, so far I've got it that cos^2 x + sin^2 x = 1. It's the next line I'm a little stumped on. I'm looking at my diagram thing and it says that cot^2 x = 1/tan x and csc^2 x = 1/sin x. Is this correct, and what do I do next?

Answer 1

\[1+\cot ^2x=\csc^2x \]

Answer 2

Your sheet sounds a smidge wrong. The csc^2x should be equal to 1/sin^2x, and cot^2x should be 1/tan^2x

Answer 3

yeah

Answer 4

@rvc that property he listed should help you simplify the equation even further.

Answer 5

further we can write tanx=sinx/cosx

Answer 6

Oh, ok. That makes sense to continue the ^2 x's across the board.

Answer 7

@rvc ok, once you do that, then what?

1 1
----------------------- -----> -------------------------------- ?
tan^2 x - 1/sin^2 x sin^2 x/cos^2x - 1/sin^2 x

Answer 8

\[\Large\rm \frac{1}{\left(\frac{\cos^2x}{\sin^2x}+\frac{1}{\sin^2x}\right)}\]You already have a common denominator down there, so combine your fractions.\[\Large\rm =\frac{1}{\left(\frac{\cos^2x+1}{\sin^2x}\right)}\]Yah? :)

Answer 9

And then flip it, apply your "keep change flip" or whatever :P

Answer 10

OH MY GOSH THAT MAKES TOTAL SENSE! Thank you!! @zepdrix

Answer 11

lol np c:

Answer 12

Alright, so to finish up here, when I flip it it gives me:

\[\frac{ \cos^2x - 1 }{ \sin^2x }\]

Right? Or am I thinking wrong?

Answer 13

Alright! Thank you all for your help! It is much appreciated!!

Answer 14

Hmmm not sure what you did there 0_o\[\Large\rm =\frac{1}{\left(\frac{\cos^2x+1}{\sin^2x}\right)}=1\cdot\left(\frac{\sin^2x}{\cos^2x+1}\right)\]I flipped that bottom fraction, ya? :)