Question

In ∆ABC, if cot A/2, cot B/2, cot C/2 are in A.P then show that a,b,c are in A.P​

Answer 1

Answer:

Step-by-step explanation:

Given :

∆ABC

$cot(\frac{A}{2}), \ cot(\frac{B}{2}), \ cot(\frac{C}{2})$ are in AP.

Procedure :

As $cot(\frac{A}{2}) = \sqrt{\frac{s(s-a)}{(s-b)(s-c)} }$

As $cot(\frac{B}{2}) = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} }$

As $cot(\frac{C}{2}) = \sqrt{\frac{s(s-c)}{(s-a)(s-b)} }$

$\sqrt{\frac{s(s-a)}{(s-b)(s-c)} }$, $\sqrt{\frac{s(s-b)}{(s-a)(s-c)} }$,$\sqrt{\frac{s(s-c)}{(s-a)(s-b)} }$ are in AP.

Multiplying All by $\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$

Hence $(\sqrt{s-a} )^{2}$, $(\sqrt{s-b} )^{2}$, $(\sqrt{s-c} )^{2}$ are in AP.

⇒ (s - a), (s - b), (s - c) are in AP.

⇒ (-a), (-b), (-c) are in AP.

∴ a, b, c are in AP.

Hence Proved.

Thanks !