I need to prove that the following is true. Thanks

(2tanx /1-tan^x)+(1/2cos^2x-1)= (cosx+sinx)/(cosx - sinx)
and thanks ...........

check your typing.
I tried 30º, the two sides are not equal, they differ by 1

oh , thank you Mr Reiny I'll tell my teacher this Question is Wrong .

not necessarily.

look at your first term
what does tan^x mean?

I want you to check what you typed here with your question.

Mr Reiny ,
I tried 30º, the two sides are equal
L.S=3.732
R.S=3.732

Yes, if the term at the bottom is tan^2 x,
you typed tan^x, and I read it as tan x

I got the proof, give me a bit of time to type it

I'm sorry Mr Reiny you are right it is tan^2x .

LS=
2tanx/(1-tan^2x) + 1/(2cos^2x -1)
= 2sinx/cosx [1/(1-sin^2x/cos^2x)] + 1/(2cos^2x -(sin^2x + cos^2x))
=2sinx/cosx [cos2x/(cos^2x - sin^2x)] + 1/(cos^2x - sin^2x) reduce to get same denominator
=2sinxcosx/(cos^2x-sin^2x) + 1/(cos^2x-sin^2x)
=(2sinxcosx + sin^2x + cos^2x)/(cos^2x-sin^2x)
= (cosx+sinx)(cosx+sinx)/[cosx+sinx)(cosx-sinx)]
=(cosx+sinx)/(cosx - sinx)
= Right Side!!!!!

thank you Mr Reiny

!!!

You're welcome! I'm glad I could help. It's important to double-check the terms and expressions when working on proofs to make sure there are no typos or errors in the calculations. Good job on finding the correct proof for the equation! If you have any more questions, feel free to ask.

You're welcome! I'm glad I could help. If you have any more questions, feel free to ask.