Question

prove this identity tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)

Answer 1

help pls.

Answer 2

\[tan(\alpha-\beta) = \frac{tan(\alpha) \pm tan(\beta)}{1 \mp tan(\alpha)tan(\beta)} \]

Answer 3

pls help me

Answer 4

That is the identiy you can use to try and prove this....

Answer 5

Ive already do that. But i cannot get the right answer

Answer 6

did you see the link?

Answer 7

help mp to answer this.

Answer 8

You could try using a substitution. Letting (x-y) = \(\alpha\) and (z-x) = \(\beta\) and prove that way. you get \(tan(\alpha) + tan(\beta\)...

Answer 9

whose mp?

Answer 10

in terms of \(\alpha\), \(\beta\), and \(\gamma\)

Answer 11

write it like abb0t says.. then take l.c.m of the three terms of l.h.s ...simplify the numerator and you will get r.h.s :)

Answer 12

ok :)

Answer 13

tan(x-y)+tan(y-z)+tan(z-x)
\[x-y=A; y-z= B; z-x= C\]
\[Tan(A)+Tan(B)+Tan(C)\]
\[A+B+C=2\pi\]
\[A+B = 2\pi-C\]
Tan(A+B) = Tan(2pi-C)
\[Tan(A+B) = \frac{ TanA+TanB }{ 1-TanATanB }\]
=\[Tan(2\pi-C) = \frac{ Tan2\pi-TanC }{ 1+Tan2piTanC }\]
Tan2pi = 0
So.
\[Tan(2\pi-C) = \frac{ 0-TanC }{ 1-0 } = -TanC\]
\[\frac{ TanA+TanB }{ 1-TanATanB } = -TanC\]
\[TanA+TanB = -TanC(1-TanATanB) =>
TanA+TanB =-TanC +TanATanBTanC\]
\[TanA+TanB+TanC = TanATanBTanC\]
A = x-y
B=y-z
C=z-x

Answer 14

why it is equal to 2pi?

Answer 15

Change your username do "iWillDoyourhomework"

Answer 16

x-y+y-z+z-x = 0
Tan(0) = Tan(2pi)

Answer 17

yes. get it thanks.

Answer 18

Actually,you could have just done
A+B+C = 0
Would have skipped a step.

Answer 19

I have no comeback @abb0t

Answer 20

good stuff @DoYourHomework

Answer 21

thx!