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Question: In a \(\triangle A B C\), if \(3 a = b + c\),then the value of \(\cot \frac { B } { 2 } \cot \frac {...

In a △ABC\triangle A B C, if 3a=b+c3 a = b + c,then the value of cot⁡B2cot⁡C2\cot \frac { B } { 2 } \cot \frac { C } { 2 } is

A

1

B

2

C

3\sqrt { 3 }

D

2\sqrt { 2 }

Answer

2

Explanation

Solution

cot⁡B2⋅cot⁡C2=s(s−b)(s−a)(s−c)⋅s(s−c)(s−a)(s−b)=ss−a\cot \frac { B } { 2 } \cdot \cot \frac { C } { 2 } = \sqrt { \frac { s ( s - b ) } { ( s - a ) ( s - c ) } } \cdot \sqrt { \frac { s ( s - c ) } { ( s - a ) ( s - b ) } } = \frac { s } { s - a }

Given 3a=b+c⇒a+b+c=4a3 a = b + c \Rightarrow a + b + c = 4 a

⇒ ∴ cot⁡B2⋅cot⁡C2=2aa=2\cot \frac { B } { 2 } \cdot \cot \frac { C } { 2 } = \frac { 2 a } { a } = 2.