Question

Trigonometric Formulas | Proof

LaTeX in comments

Answer 1

Prove that \[\tan{(s+t)}=\frac{\tan{s}-\tan{t}}{1+\tan{s}\tan{t}}\]

Answer 2

My work so far:\[\tan{(s+t)}=\frac{\sin{(s+t)}}{\cos{(s+t)}}=\frac{\sin{s}\cos{t}+\cos{s}\sin{t}}{\cos{s}\cos{t}-\sin{s}\sin{t}}\]

Answer 3

Am I going in the right direction here? o-o

Answer 4

Multiply the top and bottom by 1/(cos s cos t)

Answer 5

This? \[\frac{top}{bottom}\times\frac{\frac{1}{\cos{s}\cos{t}}}{\frac{1}{\cos{s}\cos{t}}}\]

Answer 6

Or do I multiply the whole thing by 1/[cos s cos t] o_o

Answer 7

Yes

No, because that is not multiply by 1

Answer 8

I could've said numerator and denominator but that is longer. And now you made me waste all that time saved by typing top and bottom.

Answer 9

What I said is equivalent to dividing every term by cos s cos t

Do that and then you'll have proved it

Answer 10

Actually I should have typed out everything but I am lazy and lag seems to hate me. e_o

Answer 11

Oh okay.

Answer 12

I wrote it out but I got stuck on how to simplify o_o

Answer 13

It's very easy to simplify. I don't want to type it out but mentally doing it I could see a bunch of things simplify

Answer 14

I'm on the numerator right now.\[\sin{s}\cos{t}+\cos{s}\sin{t}\times\frac{1}{\cos{s}\cos{t}}\]

Answer 15

Is it okay to cancel out the cos s?

Answer 16

Remember you need to multiply each term by it

Answer 17

\[\large \sin{s}\cos{t}\times\frac{1}{\cos{s}\cos{t}}+\cos{s}\sin{t}\times\frac{1}{\cos{s}\cos{t}}\]

Answer 18

Ohhh... I feel dumb now. Sorry

Answer 19

\[\frac{\frac{\sin{s}\cos{t}}{\cos{s}\cos{t}}+\frac{\cos{s}\sin{t}}{\cos{s}\cos{t}}}{\frac{\cos{s}\cos{t}}{\cos{s}\cos{t}}+\frac{\sin{s}\sin{t}}{\cos{s}\cos{t}}}\]

Answer 20

u have reached :)

Answer 21

So I got \[\frac{\frac{\sin{s}}{\cos{s}}+\frac{\sin{t}}{\cos{t}}}{1-\frac{sin{s}}{\cos{s}}\times\frac{\sin{t}}{\cos{t}}}\]
Sorry, missed the subtraction sign on the bottom.

I think I got it now, thank you! :)