Question

I need help verifying a trig identity.

Answer 1

Hey @satellite73 can you help me with another one

Answer 2

sure i think so

Answer 3

99% or more of this is algebra
if you replace cosine by \(a\) and sine by \(b\) the left hand side is
\[\frac{\frac{a^2}{b^2}}{\frac{1}{b}+1}\]

Answer 4

Sorry my internet was acting weird for a second there

Answer 5

Yea sorry I was going to say that but my internet is whacked out right now

Answer 6

ok i screwed it up anyway, let me start again

Answer 7

\[\frac{\frac{a^2}{b^2}}{\frac{1}{b}+1}\] no i was right hmmm

Answer 8

Not that you screwed up lol

Answer 9

But would cancel out sinx +1

Answer 10

\[\frac{\cos^2(x)}{\sin(x)+1}\]

Answer 11

i am not sure the question is correct

Answer 12

That's the way it is in my textbook.

Answer 13

lol i am an idiot let me start again

Answer 14

It's an online textbook through pearson or whatever but I copied the image from it so it should be right

Answer 15

HI!!

Answer 16

Hey misty

Answer 17

multiply top and bottom on the right by \(\sin^2(x)\) you get
\[\frac{cos^2(x)}{\sin(x)+\sin^2(x)}\] as a first step

Answer 18

is this the right or left side right im guessing

Answer 19

rewrite the top as
\[\frac{1-\sin^2(x)}{\sin(x)(1+\sin(x)}\]

Answer 20

oops i meant the left hand side
don't know my right from my left

Answer 21

then factor the numerator and cancel the common factor of \(1+\sin(x)\) and you get what you want

Answer 22

so i would end up with 1-sinx/sinx

Answer 23

yeah @misty has it

Answer 24

yes
that is what you want right?

Answer 25

yes i was just making sure thanks to you both

Answer 26

yw

Answer 27

\[\color\magenta\heartsuit\]