Question

If A,B,C are the angles of a triangle and if none of them is equal to pie by 2 then prove that i) tan A + tan B + tan C = tan A tan B tan C ii) cot A + cot B + cot B cot C + cot C cot A = 1

Answer 1

Answer:

ANSWER

Given

In ΔABC

tan(A+B+C)=π

tan(A+B+C)=tanπ

(1−tanAtanB−tanBtanC−tanAtanC)

(tanA+tanB+tanC−tanAtanBtanC)

=0

tanA+tanB+tanC=tanAtanBtanC

⟹

(tanAtanBtanC)

(tanA+tanB+tanC)

=1

⟹cotAcotB+cotBcotC+cotCcotA=1

So, tanA+tanB+tanC−tanAtanBtanC+cotAcotB+cotBcotC+cotCcotA

⟹0+1=1

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