Help in trigonometry?

Question

Prove:
tan(A+B+C) = tanA + tanB + tanC - tanA tanB tanC / 1 - tanAtanB - tanBtanC - tanCtanA

I really need help... I try expanding but it doesn't work

Answer 1

tan(a+b+c)
=[tana+tan(b+c)] / [1 - tanatan(b+c)]

= [tana + [ (tanb + tanc) / (1 - tanbtanc) ] ] / next line
[ 1 - tana[ (tanb + tanc) / (1 - tanbtanc) ]

multiply top and bottom by (1 - tanbtanc) gives

= [ tana(1 - tanbtanc) + tanb + tanc ] / nextline
[ (1 - tanbtanc) - tana[ (tanb + tanc) ]

= [tana + tanb + tanc - tanatanbtanc] / nextline
[ 1 - tanbtanc - tanatanb - tanatanc ]

as required

the end
.

Answer 2

The formula for

tan(x+y) = tanx + tany
-------------------
1 - tanx tany

using the same formula below considering

tanx = tana and
tany = tan(b+c)

L.H.S = tan(a+b+c)
=[tana + tan(b+c)] / [1 - tana tan(b+c)]

= [tana + [ (tanb + tanc) / (1 - tanbtanc) ] ] /
[ 1 - tana[(tanb + tanc) / (1 - tanbtanc)]]

multiplying top and bottom by (1 - tanb tanc)

we get,

= [ tana(1 - tanbtanc) + tanb + tanc ] /
[ (1 - tanbtanc) - tana[ (tanb + tanc) ]

= [tana + tanb + tanc - tanatanbtanc] /
[ 1 - tanbtanc - tanatanb - tanatanc ]
= R.H.S
Hence proved.

Answer 3

Let (B + C) = D
tan (A + B + C) = tan (A + D)
tan (A + D) = (tan A + tan D) / (1 - tan A tan D)

Top line (P)
tan A + tan (B + C)
tan A + (tan B + tan C) / (1 - tan B tan C )
tan A - tan A tan B tan C + tan B + tan C

Bottom Line (Q)
1 - tan A tan D
1 - tan A tan (B + C)
1 - tan A (tan B + tan C) / (1 - tan B tan C)
1 - tan B tan C - tan A tan B - tan A tan B tan C

P / Q
tan A + tan B + tan C - tan A tan B tan C
——————————————————-
1 - tan B tan C - tan A tan B - tan A tan C