Question

Let O be a point inside a triangle ABC such that
$\angle OAB=\angle OBC=\angle OCA=\omega$,
then show that
(a) $cot\omega =cotA+cotB+cotC.$
(b) $cose{c}^2\omega =cose{c}^{2}A+cose{c}^{2}B+cose{c}^{2}C.$

Answer 1

$\angle OCB=C-\omega$ and

$\angle BOC={180}^0-\omega -(C=\omega )={180}^0-C$.

Similarly $\angle AOB={180}^0-B$.

Now from $△AOB$, we have

$\frac{OB}{sin\omega }=\frac{AB}{sin({180}^0-B)}=\frac{c}{sinB}$

so that $OB=\frac{csin\omega }{sinB}$ ..........(1)

Again from $△OBC$, we get

$\frac{OB}{sin(C-\omega )}=\frac{BC}{sin({180}^0-c)}=\frac{c}{sinC}$

$\therefore OB=\frac{asin(C-\omega )}{sinC}$ ........(2)

From (1) and (2), we get

$\frac{csin\omega }{sinB}=\frac{asin(C-\omega )}{sinC}$

or $KsinCsin\omega sinC=ksinAsinBsin(C-\omega )$

or $sinCsin\omega sin(A+B)=sinAsinBsin(C-\omega )$

or $sinCsin\omega sinAcosB+sinCsin\omega cosAsinB$

=$sinasinBsinCcos\omega -sinAsinBsinCsin\omega$

Dividing by $sinAsinBsinCsin\omega$, we get

$cotB+cotA=cot\omega -cotC$

or $cot\omega =cotA+cotB+cotC$

Squaring,

$co{t}^2\omega =co{t}^{2}A+co{t}^{2}B+co{t}^{2}C+2cotAcotB+2cotBcotC+2cotCcotA$

or $cose{c}^2\omega -1=cose{c}^{2}A-cose{c}^{2}B-1+cose{c}^{2}C-1+2$

[$\because$ in a $△$, cot A cotB+ cot B cot C + cot C cot A =1]

or $cose{c}^2\omega =cose{c}^{2}A+cose{c}^{2}B+cose{c}^{2}C.$