if a+b+c=2pie then find the value of tana/2+tanb/2+tanc/2 is equal to
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Tan(a/2)Tan(b/2)Tan(c/2)
(a+b+c) /2=2 pi
(a+b)/2=pi-c/2
So applying tan on both sides we get
{Tana/2 +tanb/2}/1-tana/2*tanb/2 =tan(pi-c/2)
Since tan(pi -x)= (-tanx)
And cross multiply we get
Tana/2+tanb/2+tanc/2=tana/2*tanb/2*tanc/2
(a+b+c) /2=2 pi
(a+b)/2=pi-c/2
So applying tan on both sides we get
{Tana/2 +tanb/2}/1-tana/2*tanb/2 =tan(pi-c/2)
Since tan(pi -x)= (-tanx)
And cross multiply we get
Tana/2+tanb/2+tanc/2=tana/2*tanb/2*tanc/2
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