Question

In a triangle $ABC$, prove that
(i) $\cot \frac{A}{2}-\cot \frac{B}{2}=\frac{(b-a)s}{\Delta }$
(ii) $(b+c)\sin \frac{A}{2}=a\cos \frac{B-C}{2}$
(iii) $bc{\cos }^2\frac{A}{2}+ca{\cos }^2\frac{B}{2}+ab{\cos }^2\frac{C}{2}=s^2$

Answer 1

(i) $\cot \frac{A}{2}-\cot \frac{B}{2}=\frac{s(s-a)}{\Delta }-\frac{s(s-b)}{\Delta }=\frac{s}{\Delta }(s-a-s+b)=\frac{s(b-a)}{\Delta }$

(ii) As $\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

$\Rightarrow (b+c)\sin \frac{A}{2}=2R(\sin B+\sin C)\sin \frac{A}{2}$

$=2R\left(2\sin \left(\frac{B+C}{2}\right)\cos \left(\frac{B-C}{2}\right)\right)\sin \frac{A}{2}=4R\cos \frac{A}{2}\cos \left(\frac{B-C}{2}\right)\sin \frac{A}{2}=2R\sin A\times \cos \left(\frac{B-C}{2}\right)$

$=a\cos \left(\frac{B-C}{2}\right)$

(iii) $bc{\cos }^2\frac{A}{2}=bc\times \frac{s(s-a)}{bc}$

$\Rightarrow s(s-a+s-b+s-c)$

$\Rightarrow s(3s-2s)=s^2$.