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# C3 Trig Identities HELP Watch

1. Hi,
I'm stuck on a c3 trig question. Was wondering if any of you may be able to give some help.

(Q) Prove the following identity:

tan(a+45) + tan(a-45) = 2tan2a

ayathullah
2. (Original post by ayathullah)
Hi,
I'm stuck on a c3 trig question. Was wondering if any of you may be able to give some help.

(Q) Prove the following identity:

tan(a+45) + tan(a-45) = 2tan2a

ayathullah
.
Use this to expand the LHS and then manipulate it until you have the RHS.
Note that .
3. Hi,

Just a little bit confused...........

(Original post by JamieJ_93)

= [- (tanA-1)^2 +(tanA+1)^2 ]/(1-tan^2A)
Any chance you could tell me how you multiplied out the equation above.

(Original post by JamieJ_93)

= (2tanA + 2tanA ) / (1-tan^2A)
And how did you get rid of the 1-tan^2A from the bottom to get to 2tan(A+A)

Thanks,
ayathullah
4. right, so first i can see already got , so using the identities for tan(a-b) and tan(a+b) with this..

you end up with:

[Tan(a) +1 ] / [1-Tan(a)] - [Tan(a) +1] / [1+ Tan(a)]

then you combine this into one fraction to give

[Tan(a) +Tan(a) +1+ [Tan(a)]squared ) + (Tan(a) +Tan(a) -1 - [Tan(a)]squared]

all over: [1-[Tan(a)]squared -Tan(a) +Tan(a)]

simplifying this you get:

[4Tan(a)] / [1-[Tan(a)]squared]

=2[2Tan(a)] / [1-[Tan(a)]squared]

( using the double angle identity: Tan(2a) = [2Tan(a)]/(1-[Tan(a)]squared)

=2Tan(2a)

hope you understood that, i have yet to master tex
5. i understand it now many thnaks for all your help

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