Question

@dude @angle @ people who can help me xd

Answer 1

I think ultri was tempted to make an @ everyone tag at some point that only mods could use

Answer 2

LOL

Answer 3

well, i did this so far., i kinda expanded, not sure what i was supposed to do lol
kinda that

Answer 4

this?
\(\huge\frac{\frac{cos^2x-sin^2x}{cos^2x}}{\frac{cos^2x+sin^2x}{cos^2x}}\)

Answer 5

Yea, that lol

Answer 6

multiply top and bottom by \(cos^2x\)

Answer 7

those cross out?
so then
(cos^2 x - sin^2 x)/(cos^2 x + sin^2 x)

Answer 8

yes

then what can you replace the denominator with?

Answer 9

1?

Answer 10

so then
cos^2 x - sin^2 x

Answer 11

yes yes

have you learned about cos(2x)?

Answer 12

No?
actually, well, we don't learn anything in class?

Answer 13

LOL

Answer 14

eh, I guess they might want us to go the long way then

Answer 15

so it's cos 2x?

Answer 16

nono backtrack a bit
cos(2x) is a shortcut, but if you haven't learned it then we're not gonna do that

Answer 17

\(\color{#0cbb34}{\text{Originally Posted by}}\) snowflake0531
so then
cos^2 x - sin^2 x
\(\color{#0cbb34}{\text{End of Quote}}\)

replace cos^2(x) = 1 - sin^2(x)

Answer 18

oh, right
and there's my answer
LOL
thanksssssssssssssssssssssssssssss xdxdxdxdxd <3

Answer 19

:)

Answer 20

<3

Answer 21

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

Answer 22

thanks @surjithayer !

Answer 23

yw

Answer 24

thanks~~!