Question

in a ∆ABC if 3a=b+c then find value of CotB/2 .Cot C/2

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Answer 1

$\bf{\Huge{\boxed{\rm{\blue{ANSWER\::}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

In a ΔABC, If 3a = b + c.

$\bf{\Large{\underline{\bf{To\:find\::}}}}$

The value of cotB/2 * cotC/2.

$\bf{\Large{\underline{\tt{\red{Explanation\::}}}}}$

We know that formula of the semi - perimeter of Δ:

$\leadsto\sf{\Large{\boxed{\rm{Semi\:-\:perimeter\:(S)=\:\frac{a+b+c}{2} }}}}}$

A/q

$\longmapsto\sf{S\:=\:\frac{a+b+c}{2} }$

$\longmapsto\sf{S\:=\:\frac{a+3a}{2} }$

$\longmapsto\sf{S\:=\:\cancel{\frac{4a}{2}} }$

$\longmapsto\sf{\pink{S\:=\:2a}}$

Now,

$\longmapsto\sf{cot\frac{B}{2} .cot\frac{C}{2} \:=\:\sqrt{\frac{(s)(s-b)}{(s-a)(s-c)} } .\sqrt{\frac{s(s-c)}{(s-a)(s-b)} } }$

$\longmapsto\sf{cot\frac{B}{2} .cot\frac{C}{2} \:=\:\sqrt{\frac{s^{2} (s-b)(s-c)}{(s-a)^{2} (s-b)(s-c)} } }$

$\longmapsto\sf{cot\frac{B}{2} .cot\frac{C}{2} \:=\:\sqrt{\frac{s^{2} \cancel{(s-b)(s-c)}}{(s-a)^{2} \cancel{(s-b)(s-c)}} } }$

$\longmapsto\sf{cot\frac{B}{2} .cot\frac{C}{2} \:=\:\frac{s}{s-a} }$

$\longmapsto\sf{cot\frac{B}{2} .cot\frac{C}{2} \:=\:\frac{2a}{2a-a} }$

$\longmapsto\sf{cot\frac{B}{2} .cot\frac{C}{2} \:=\:\cancel{\frac{2a}{a} }}$

$\longmapsto\sf{\red{cot\frac{B}{2} .cot\frac{C}{2} \:=\:2} }$

Thus,

$\bf{\Large{\boxed{\tt{The\:\:value\:\:is\:\:\:\:2}}}}}$