Question

Simplify the expression: 1/1+cot^2x
sec2x
csc2x
sin2x
cos2x
tan2x

Answer 1

@AloneS.

Answer 2

I guess the problem is 1/(1+cot^2(x)) ?

Answer 3

or in latex
\[\frac{1}{1+\cot^2(x)}\]
use a Pythagorean identity for the denominator

Answer 4

i dont understand how you would use it im so lost

Answer 5

\[\sin^2(x)+\cos^2(x)=1 \\ \tan^2(x)+1=\sec^2(x) \\ 1+\cot^2(x)=\csc^2(x)\]

these are the Pythagorean identities

Answer 6

you only need one of these here

Answer 7

can you make a guess to which one

Answer 8

the last one because it has what your working with in it ?

Answer 9

right
so are you saying you don't know how to use it here for this problem?

Answer 10

no i tried and got something that was not there

Answer 11

what did you get

Answer 12

cot^2 which idk how i got it tho

Answer 13

I don't know how you got that :p

Answer 14

you do know that by the Pythagorean identity we mentioned above that you can replace 1+cot^2(x) with csc^2(x) since 1+cot^2(x)=csc^2(x) ?

Answer 15

?

Answer 16

do you see the 1+cot^2(x) on bottom

do you know why I mentioned the Pythagorean identity 1+cot^2(x)=csc^2(x)

it is because you have 1+cot^2(x) on bottom

you can replace 1+cot^2(x) with csc^2(x)

since 1+cot^2(x)= csc^2(x)

Answer 17

1 + tan^2(x)
___________
1 + cot^2(x)

1 + tan^2(x)
_____________
1 + (1 / tan^2(x))

1 + tan^2(x)
___________________
(1 + tan^2(x))/tan^2(x)

= tan^2(x)

b)

sec^2(x) cot^2(x)

= (1/cos^2(x))(cos^2(x)/sin^2(x))

= 1/sin^2(x)

= csc^2(x)

Answer 18

This is what i got.

Answer 19

so its d then since i finally think i got the work right

Answer 20

can I see what you think we have after using the Pythagorean identity I mentioned
don't do anything else
just use the Pythagorean identity I mentioned above

Answer 21

brb class switch

Answer 22

k