Question

In triangle ABC if b+c=3a,then prove that cot B/2 cot C/2 = 2

Answer 1

Answer:

Given : b+c = 3a

According to the half range formula, $Cot\frac { B }{ 2 } =\sqrt { \frac { (s-a)(s-c) }{ s(s-b) } }$

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

We know that ${\frac{(a+b+c)}{2}} = s$, given that b + c = 3a

                $S = {\frac{( a + 3a )}{2}} ={\frac{ 4a}{ 2}} = 2a$

$Cot\frac { B }{ 2 } Cot\frac { C }{ 2 } = \sqrt { \frac { (s-a)(s-c) }{s(s-b) } } \times \sqrt { \frac { (s-a)(s-b) }{ s(s-c) } }$

              $= \sqrt { \frac { (2a-a)(2a-c) }{ 2a(2a-b) } } \times \quad \sqrt { \frac { (2a-a)(2a-b) }{ 2a(2a-c) } }$

               = 2

Therefore, $Cot\frac { B }{ 2 } Cot\frac { C }{ 2 } = 2$

Answer 2

Answer:

step by step explanation hope it helps you