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In triangle ABC if \tan \frac{A}{2},\tan \frac{B}{2},\tan \frac{C}{2} are in H.P. then sides of triangle ABC will be in ?

Option: 1

A.P.


Option: 2

G.P.


Option: 3

H.P.


Option: 4

None of these


Answers (1)

best_answer

 

 

Half-Angle Formula (in terms of perimeter and sides of triangle)(part 2) -

Half-Angle Formula (in terms of perimeter and sides of triangle)(part 2)

\\\mathrm{\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}}\\\\\mathrm{\tan\frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}}\\\\\mathrm{\tan\frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}}

 

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\tan \frac{A}{2},\tan \frac{B}{2},\tan \frac{C}{2} \ are\ in\ H.P.\rightarrow \frac{2}{\tan \frac{B}{2}}=\frac{1}{\tan \frac{A}{2}}+\frac{1}{\tan \frac{C}{2}}\\ 2 \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}+\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}\\ 2 \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}=\sqrt{\frac{s(s-a)^2+s(s-b)^2}{(s-a)(s-b)(s-c)}}\\ 2 \sqrt{s(s-b)^2}=\sqrt{s[(s-a)^2+(s-b)^2]}\\ 2(s-b)=(s-a)+(s-c)\\ a+c=2b\\ a,b,c \ is\ in\ A.P.

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Gautam harsolia

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