Question

if A+B+C=pai prove that tan2A+tan2B+tan2C=tan2A.tan2B.tan2C

Answer 1

Hey mate your answer is her
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Given,
A+B+C=pai

proof that,
tanA+tanB+tanC=tanA•tanB•tanC
solutions,
=>2A+2B+2C=2pai-2C
=>tan(2A+2B)=tan(2pai-2C)
=>tan2A+tan2B/1-tan2Atan2B= -tan2C
=>tan(2A+2B+2C)=tan(2A•2B•2C)
-------------------------------------------- proved-------




hope this answers is helpful.....

Answer 2

Answer:  The proof is done below.

Step-by-step explanation:  We are given that :

$A+B+C=\pi~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We are to prove that

$\tan2A+\tan2B+\tan2C=\tan2A\tan2B\tan2C.$

We ill be using the following formula :

$(i)~\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y},\\\\\\(ii)~\tan(2\pi-x)=-\tan x.$

From equation (i), we have

$A+B+C=\pi\\\\\Rightarrow 2(A+B+C)=2\pi~~~~~~~~~~~~~~~~~~~[\textup{multiplying both sides by 2}]\\\\\Rightarrow2A+2B=2\pi-2C\\\\\Rightarrow \tan(2A+2B)=\tan(2\pi-2C)\\\\\\\Rightarrow \dfrac{\tan2A+\tan2B}{1-\tan2A\tan2B}=-\tan2C\\\\\Rightarrow \tan2A+\tan2B=-\tan2C(1-\tan2A\tan2B)\\\\\Rightarrow \tan2A+\tan2B=-\tan2C+\tan2A\tan2B\tan2C\\\\\Rightarrow \tan2A+\tan2B+\tan2C=\tan2A\tan2B\tan2C.$

Hence proved.