Question

prove/solve tan4t- tan2t/1+tan4t tan2t = 2tant/1-tan^2t

Answer 1

Too bad that cannot be rationally translated. Please remember your Order of Operations and add parentheses where necessary.

Examples:

tan(x) + 1/sin(x) = \(\tan{x} + \dfrac{1}{\sin(x)}\)

(tan(x) + 1)/sin(x) = \(\dfrac{\tan(x) + 1}{\sin(x)}\)

1/tan(x) + 2 = \(\dfrac{1}{\tan(x)} + 2\)

1/(tan(x) + 2) = \(\dfrac{1}{\tan(x) + 2}\)

Answer 2

(tan4T- tan2T)/(1+tan4T tan2T) = 2tanT/1-tan^2T

Answer 3

Are you sure? How about that last piece on the far right? Words and symbols mean things.

Answer 4

yep, i mean you can put both top and bottom of the right side in parentheses regardless but your supposed to use the trig identities to prove it

Answer 5

It does help to write it correctly, first.

Do you have this? \(\tan(\alpha - \beta) = \dfrac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}\). It looks like it could be useful on the left-hand side.

And this? \(\tan(\alpha + \beta) = \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}\). It looks like it could be useful on the right-hand side.

Answer 6

yea sorry about that the interface is new to me. i got tan(4T-2T) for the left side im not sure if that is correct and if it is what to do next

Answer 7

Perfect.

\(\tan(4T - 2T) = \tan(2T) = \tan(T + T)\)

Now, use the other formula! You'll be done!

Answer 8

awesome! thanks for your help/support!

Answer 9

You got it!! Good work.