Question

verifying trig function
need help getting started

sec(a+b) = sec(a)sec(b)/1-tan(a)tan(b)

Answer 1

@calculusfunctions

Answer 2

Let's start with the left side again. You know that

cos (A + B) = (cos A)(cos B) - (sin A)(sin B).

Correct @biochems2k ?

Answer 3

yeah would it be 1/(cos A)(cos B) - (sin A)(sin B) because its secant?

Answer 4

Yes!! Now keeping in mind that we have to manipulate this left side to get the right side, observe closely, the right side. You see that the right side has tangent. You know that tangent is sine/cosine. Thus divide both the numerator and the denominator by
(cos A)(cos B). Go ahead and do that.

Answer 5

im not sure if i did it correctly but at the end would i have 1/1-(sin a sin b)?

Answer 6

because i changed the right side to have everything in sine and cosine

Answer 7

No, Allow me to show you. Give me a couple of minutes to type it up.

Answer 8

\[L.S.=\sec (A +B)\]
\[=\frac{ 1 }{ \cos (A +B) }\]
\[=\frac{ 1 }{ \cos A \cos B -\sin A \sin B }\]
\[=\frac{ \frac{ 1 }{ \cos A \cos B } }{ \frac{ \cos A \cos B }{ \cos A \cos B }-\frac{ \sin A \sin B }{ \cos A \cos B } }\]
Do you now see what I meant? Now can you continue from there @biochems2k ?

Answer 9

@biochems2k do you not see what happens when you simplify from where I left Off?

Answer 10

I have to go @biochems2k I would like to know that you understand before I go. Could you please say something? I'm waiting.

Answer 11

@calculusfunctions im stumped where you got large fraction towards the end

Answer 12

I divided each and every term of the numerator and the denominator by (cos A)(cos B)

Answer 13

what happens to the top fraction?

Answer 14

Now the numerator is (sec A)(sec B) and the denominator is 1 - (tan A)(tan B) Don't you see?

Answer 15

oh yes! i see it now. i just saw one was cos a and b

thanks!

Answer 16

Are you sure you understand?

Answer 17

Is it alright if I go? Do you need anything else?

Answer 18

yeah i think i can finish it. i dont want you to give me the answer. just lead me in the correct way. thanks a bunch

Answer 19

Welcome! I'll probably be back in a few hours if you need anything else. Bye!!