Question

1 last trig problem

Answer 1

Try rewriting everything in terms of either sine or cosine

Answer 2

eg: sec(x) = 1/cos(x)

Answer 3

ahhhhh ok

Answer 4

give me one sec

Answer 5

also, when it comes to proving identities, you should only work on one side

Answer 6

it might be easier to work on the right side

Answer 7

\[q \cot q/\cos^{-1} -\sin(\theta)/expsec/\csc ^2\frac{ f1/\sec}{ ? } \]

Answer 8

CRAP, i hate uploading equations on this website

Answer 9

one sec, let me try again

Answer 10

ok I got my answer, but when I try to put it in the equation on this website it messes up

Answer 11

so i started working on the right side

Answer 12

After trying out the right side, it's somewhat more complicated than I thought. So I'm going to work on the left side instead
\[\Large \frac{\sec(\beta)}{\sin(\beta)}-\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\]
\[\Large {\color{red}{\frac{\sec(\beta)}{\sec(\beta)}*}}\frac{\sec(\beta)}{\sin(\beta)}-\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\]
\[\Large \frac{\sec^2(\beta)}{\sin(\beta)\sec(\beta)}-\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\]
\[\Large \frac{\sec^2(\beta)}{\sin(\beta)\sec(\beta)}-{\color{red}{\frac{\sin(\beta)}{\sin(\beta)}}}*\frac{\sin(\beta)}{\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\]
\[\Large \frac{\sec^2(\beta)}{\sin(\beta)\sec(\beta)}-\frac{\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\]
\[\Large \frac{\sec^2(\beta)-\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\]
Take a few minutes and look all that over. Then let me know when you're ready for the next step.

Answer 13

wait I think I need to prove the right side of the equation not the left

Answer 14

you can start on either side

Answer 15

ok i get it

Answer 16

just as long as you only work with one side

Answer 17

alright is that it

Answer 18

@jim_thompson5910

Answer 19

look at the "Pythagorean Identities" on page 2

Answer 20

what is sec^2 equal to?

Answer 21

tan^2(theta)+1

Answer 22

and what is sin^2 equal to?

Answer 23

1-cos^2(theta)

Answer 24

So,
\[\Large \frac{\sec^2(\beta)-\sin^2(\beta)}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\]
\[\Large \frac{\tan^2(\beta)+1-(1-\cos^2(\beta))}{\sin(\beta)\sec(\beta)} = \frac{\tan^2(\beta)+\cos^2(\beta)}{\tan(\beta)}\]

Answer 25

simplify the numerator

for the denominator, use the fact that sin*sec = sin*(1/cos) = sin/cos = ???

Answer 26

Ok I did it, can you show me what you got

Answer 27

when you simplified the numerator, what did you get?

Answer 28

i wish I could show you, trust me, but I cant

Answer 29

draw it out

Answer 30

or type what you can

Answer 31

I got back to where we started,

Answer 32

\[\Large \tan^2(\beta)+1-(1-\cos^2(\beta))\]

\[\Large \tan^2(\beta)+1-1+\cos^2(\beta)\]

which becomes what?

Answer 33

what the other side of the equation is

Answer 34

yeah ultimately the goal is to get the left side equal to the right side

Answer 35

what about the denominator

Answer 36

what is sin(x)*sec(x) equal to?

Answer 37

hint:
look at "Tangent and Cotangent Identities" and "Reciprocal Identities" on that pdf

Answer 38

ok I got it

Answer 39

it is sinxsec

Answer 40

sin(x)*sec(x) = sin(x)*(1/cos(x)) = sin(x)/cos(x) = ????

Answer 41

tan

Answer 42

yep

Answer 43

alright thanks