In a - ABC- If tanA2-tanB2-tanC2 are in H-P- then a- b- c are in-

TanA2,tanB2,tanC2 are in H.P {∵tanA2=Δs(s b) nbsp; nbsp; } ⇒Δs(s a),Δs(s b),Δs(s c) are in H.P So,s(s a)Δ, nbsp;s(s b)Δ,s(s c)Δ are in A.P ⇒ 2×s(s b)Δ = s ...

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Standard XII

Mathematics

Semi Perimeter

In a Δ ABC. I...

Question

In a $Δ A B C$ . If $tan \frac{A}{2}, tan \frac{B}{2}, tan \frac{C}{2}$ are in $H . P$ , then $a, b, c$ are in:

H . P

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G . P

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A . P

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A . G . P

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Solution

The correct option is C $A . P$
$tan \frac{A}{2}, tan \frac{B}{2}, tan \frac{C}{2}$ are in $H . P$
${∵ tan \frac{A}{2} = \frac{Δ}{s (s - b)}}$
$\Rightarrow \frac{Δ}{s (s - a)}, \frac{Δ}{s (s - b)}, \frac{Δ}{s (s - c)} are in H.P$
So, $\frac{s (s - a)}{Δ}, \frac{s (s - b)}{Δ}, \frac{s (s - c)}{Δ} are in A.P$
$\Rightarrow 2 \times \frac{s (s - b)}{Δ} = \frac{s (s - a)}{Δ} + \frac{s (s - c)}{Δ}$
$\Rightarrow 2 (s - b) = (s - a) + (s - c)$
$\Rightarrow 2 s - 2 b = 2 s - a - c$
$\Rightarrow 2 b = a + c$
$\Rightarrow a, b, c are in A.P$

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In a ΔABC. If tanA2,tanB2,tanC2 are in H.P, then a,b,c are in:

The correct option is C A.PtanA2,tanB2,tanC2 are in H.P {∵tanA2=Δs(s−b)} ⇒Δs(s−a),Δs(s−b),Δs(s−c) are in H.P So, s(s−a)Δ,s(s−b)Δ,s(s−c)Δ are in A.P ⇒2×s(s−b)Δ=s(s−a)Δ+s(s−c)Δ ⇒2(s−b)=(s−a)+(s−c) ⇒2s−2b=2s−a−c ⇒2b=a+c ⇒a, b, c are in A.P

In a $Δ A B C$ . If $tan \frac{A}{2}, tan \frac{B}{2}, tan \frac{C}{2}$ are in $H . P$ , then $a, b, c$ are in: