Question

If A,B,C are angles of a triangle and none of them is equal to π/2, then prove that. tan A + tanB + tanc = tan A. tanB. tanC.​

Answer 1

Answer:

Step-by-step explanation:

Given: tanA+tanB+tanC=tanAtanBtanC

⇒tanA+tanB=tanAtanBtanC−tanC

⇒tanA+tanB=tanC(tanAtanB−1)

⇒

1−tanAtanB

tanA+tanB

​

=−tanC

⇒tan(A+B)=−tanC

⇒A+B=−C⇒A+B+C=0

or some integral multiple of π

Answer 2

SOLUTION

GIVEN

A, B, C are angles of triangle and none of them equal to π/2

TO PROVE

tan A + tan B +tan C = tanA.tanB.tanC

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf{ \tan(A + B) = \frac{ \tan A + \tan B}{1 - \tan A \tan B} }$

EVALUATION

Here it is given that

$\displaystyle \sf{A + B + C = \pi}$

$\displaystyle \sf{ \implies \: A + B = \pi - C}$

$\displaystyle \sf{ \implies \tan(A + B) = \tan( \pi - C)}$

$\displaystyle \sf{ \implies \tan(A + B) = - \tan C}$

$\displaystyle \sf{ \implies \frac{ \tan A + \tan B}{1 - \tan A \tan B} = - \tan C}$

$\displaystyle \sf{ \implies \tan A + \tan B = \: - \tan C + \tan A \tan B \tan C}$

$\displaystyle \sf{ \implies \tan A + \tan B + \tan C = \tan A \tan B \tan C}$

Hence proved

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