Question

Csc^2 theta = 3Sec^2 theta.

How do convert either Csc to sec or Sec to csc? Also how do you solve it in radians?

Answer 1

i would start with
\[\sin^2(\theta)=\frac{\cos^2(\theta)}{3}\] by taking reciprocals

Answer 2

or maybe go right to
\[\tan^2(\theta)=\frac{1}{3}\]

Answer 3

Sorry I don't really understand how you got to that response :/
I'm new to this.

Answer 4

ok i can't work with cosecant , secant etc
lets go step by step and work with since and cosine

Answer 5

since \(\csc(x)=\frac{1}{\sin(x)}\) and \(\sec(x)=\frac{1}{\cos(x)}\) your original equation is
\[\frac{1}{\sin^2(x)}=\frac{3}{\cos^2(x)}\]

Answer 6

solve as any ratio first, get \[\cos^2(x)=3\sin^2(x)\]

Answer 7

so far so good? only algebra at these steps no trig

Answer 8

yeah, i didn't thing about changing back to sin and cos

Answer 9

eventually you should come to
\[\tan^2(x)=\frac{1}{3}\] making
\[\tan(x)=\pm\frac{1}{\sqrt3}\]

Answer 10

cos/3sin right?

Answer 11

gives tan?

Answer 12

I would be careful about dividing both sides by 0

\[\cos^2(x)=3\sin^2(x) \\ 1-\sin^2(x)=3\sin^2(x) \\ 1=4\sin^2(x) \]

Answer 13

but given the initial problem sin(x) cannot be 0 and neither can cosine
so you are good

Answer 14

Yeah I know that cos^2 = 1-sin^2 but I didn't know you couldnt go it with it csc and sec

Answer 15

sat's way will work
i'm just saying we need to be careful when we divide by 0
but really we aren't dividing by 0 because we had csc and sec in the initial problem

Answer 16

Oh okay, I got you, Thanks :)

Answer 17

I have found my answer and really appreciate your guy's help!
It really means a lot to me :)

Answer 18

congrats