Question

prove that cotA/2+cotB/2+cotC/2=s2/∆​

Answer 1

Given Question

Prove that

$\rm :\longmapsto\:cot\dfrac{A}{2} + cot\dfrac{B}{2} + cot\dfrac{C}{2} = \dfrac{ {s}^{2} }{\triangle }$

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

Consider, LHS

$\rm :\longmapsto\:cot\dfrac{A}{2} + cot\dfrac{B}{2} + cot\dfrac{C}{2}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ tan \frac{A}{2} = \frac{r}{s - a} \: }}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{s - a}{r} + \dfrac{s - b}{r} + \dfrac{s - c}{r}$

On taking LCM, we get

$\rm \:  =  \: \dfrac{s - a + s - b + s - c}{r}$

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

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ a + b + c = 2s \: }}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{3s - 2s}{r}$

$\rm \:  =  \: \dfrac{s}{r}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ r = \frac{\triangle }{s}}}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{s}{ \: \: \: \dfrac{\triangle}{s} \: \: \: }$

$\rm \:  =  \: \dfrac{ {s}^{2} }{\triangle }$

Hence,

$\rm\implies \:\:\boxed{\tt{ cot\dfrac{A}{2} + cot\dfrac{B}{2} + cot\dfrac{C}{2} = \dfrac{ {s}^{2} }{\triangle } }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

LEARN MORE

$\boxed{\tt{ sin \frac{A}{2} = \sqrt{ \frac{(s - b)(s - c)}{bc} } \: }}$

$\boxed{\tt{ sin \frac{B}{2} = \sqrt{ \frac{(s - a)(s - c)}{ac} } \: }}$

$\boxed{\tt{ sin \frac{C}{2} = \sqrt{ \frac{(s - a)(s - b)}{ab} } \: }}$

$\boxed{\tt{ cos\frac{A}{2} = \sqrt{ \frac{s(s - a)}{bc} } \: }}$

$\boxed{\tt{ cos\frac{B}{2} = \sqrt{ \frac{s(s - b)}{ca} } \: }}$

$\boxed{\tt{ cos\frac{C}{2} = \sqrt{ \frac{s(s - c)}{ab} } \: }}$

$\boxed{\tt{tan\frac{A}{2} = \sqrt{ \frac{(s - b)(s - c)}{s(s - a)} } \: }}$

$\boxed{\tt{tan\frac{B}{2} = \sqrt{ \frac{(s - a)(s - c)}{s(s - b)} } \: }}$

$\boxed{\tt{tan\frac{C}{2} = \sqrt{ \frac{(s - a)(s - b)}{s(s - c)} } \: }}$