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i tried left side: 1+tanx+1-tanx
=2 :S

have you studied double angle identities?

well i have the formulas for it

ok good tell me formula for tan(a+b) and tan (a-b)
remember a and b are angles.

tan(a+/-b)=(tana+/-tanb)/(1+/-tanatanb)

and then do i conjugate?

i ended up with tanx

actually it was 1+tan^2x/1+tanx

hmm i dont think so you did right.. as far as i know there should be
1-tan^2x in the denominator.

i got (1+tanx)(1-tanx)in deno

yes.. multiplying it would give (1-tan^2x)

right?

(tanpi/4+tanx)^2 + (tanpi/4-tanx) / (1-tanpi/4tanx)(1+tanpi/4tanx)

and then tanpi/4 =1

2+tan^2x-tanx / 1-tan^2x

ru there?

yes i am.. just solving for you..

i multiplied (tan pi/4 +tanx) to both top and bottom

it will become a common deno

tell me if this helps you.