Question

Suppose x = tan A, y = tan B, z = tan C.

Suppose none of A,B,C, A-B, B-C, C-A is an odd multiple of
2
x- y
Then prove that
Σ(x-y/1+xy)
= II
(1 + xy
1 + xy)​

Answer 1

Answer:

Correct option is

C

nπ

Given x=tanA,y=tanB,z=tanC

and x+y+z−xyz=0

∴tan(A+B+C)=

1−(tanAtanB+tanBtanC+tanCtanA)

tanA+tanB+tanC−tanAtanBtanC

tan(A+B+C)=

1−(xy+yz+xz)

x+y+z−xyz

=0

⟹tan(A+B+C)=0

⟹(A+B+C)=tan

−1

(0)=nπ

⟹A+B+C=nπ

Step-by-step explanation:

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