Question

Hey guys, please help me with this question (class 12th , chapter 3 : Trigonometric Functions)​

Answer 1

Given Question :-

Prove that, In triangle ABC

$\rm \: cot\bigg(\dfrac{A}{2} \bigg) + cot\bigg(\dfrac{B}{2} \bigg) + cot\bigg(\dfrac{C}{2} \bigg) = \dfrac{a + b + c}{b + c - a}cot\bigg(\dfrac{A}{2} \bigg)$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: cot\bigg(\dfrac{A}{2} \bigg) + cot\bigg(\dfrac{B}{2} \bigg) + cot\bigg(\dfrac{C}{2} \bigg)$

We know,

$\boxed{\sf{  \:cot\bigg(\dfrac{A}{2} \bigg) = \sqrt{ \frac{s(s - a)}{(s - b)(s - c)} } \: }}$

So, using this result, we get

$\rm \: =  \: \sqrt{\dfrac{s(s - a)}{(s - b)(s - c)} } + \sqrt{\dfrac{s(s - b)}{(s - c)(s - a)} } + \sqrt{\dfrac{s(s - c)}{(s - a)(s - b)} }$

can be rewritten as

$\rm \: = \: \sqrt{\dfrac{s(s - a)^{2} }{(s - a)(s - b)(s - c)} } + \sqrt{\dfrac{s(s - b) ^{2} }{(s - c)(s - a)(s - b)} } + \sqrt{\dfrac{s(s - c)^{2} }{(s - a)(s - b)(s - c)} }$

$\rm \: =  \: \sqrt{\dfrac{s}{(s - a)(s - b)(s - c)} }(s - a + s - b + s - c)$

$\rm \: =  \: \sqrt{\dfrac{s}{(s - a)(s - b)(s - c)} }[3s - (a + b + c)]$

$\rm \: =  \: \sqrt{\dfrac{s}{(s - a)(s - b)(s - c)} }[3s - 2s]$

$\rm \: =  \: s \: \sqrt{\dfrac{s}{(s - a)(s - b)(s - c)} }$

can be further rewritten as

$\rm \: =  \: s \: \sqrt{\dfrac{s(s - a)}{(s - a)^{2} (s - b)(s - c)} }$

can be further rewritten as

$\rm \: =  \: \dfrac{s}{s - a} \: \sqrt{\dfrac{s(s - a)}{(s - b)(s - c)} }$

can be further rewritten as

$\rm \: =  \: \dfrac{2s}{2(s - a)} \: cot\bigg(\dfrac{A}{2} \bigg)$

$\rm \: =  \: \dfrac{a + b + c}{2s - 2a} \: cot\bigg(\dfrac{A}{2} \bigg)$

$\rm \: =  \: \dfrac{a + b + c}{a + b + c- 2a} \: cot\bigg(\dfrac{A}{2} \bigg)$

$\rm \: =  \: \dfrac{a + b + c}{b + c- a} \: cot\bigg(\dfrac{A}{2} \bigg)$

Hence,

$\boxed{\sf{  \:\rm \: cot\bigg(\dfrac{A}{2} \bigg) + cot\bigg(\dfrac{B}{2} \bigg) + cot\bigg(\dfrac{C}{2} \bigg) = \dfrac{a + b + c}{b + c - a}cot\bigg(\dfrac{A}{2} \bigg)}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Formulae Used :-

In triangle ABC, if the sides are represented as a, b, c respectively then perimeter of triangle ABC is

$\boxed{\sf{  \:2s = a + b + c \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information

$\rm \: sin\bigg(\dfrac{A}{2} \bigg) = \sqrt{\dfrac{(s - b)(s - c)}{bc} }$

$\rm \: sin\bigg(\dfrac{B}{2} \bigg) = \sqrt{\dfrac{(s - c)(s - a)}{ca} }$

$\rm \: sin\bigg(\dfrac{C}{2} \bigg) = \sqrt{\dfrac{(s - a)(s - b)}{ab} }$

$\rm \: cos\bigg(\dfrac{C}{2} \bigg) = \sqrt{\dfrac{s(s - c)}{ab} }$

$\rm \: cos\bigg(\dfrac{B}{2} \bigg) = \sqrt{\dfrac{s(s - b)}{ac} }$

$\rm \: cos\bigg(\dfrac{A}{2} \bigg) = \sqrt{\dfrac{s(s - a)}{bc} }$

$\rm \: tan\bigg(\dfrac{A}{2} \bigg) = \sqrt{\dfrac{(s - b)(s - c)}{s(s - a)} }$