Question

Prove that:

For any $△ABC,$ $a\cos B\cos C+b\cos C\cos A+c\cos A\cos B=\frac{\Delta }{R}$ where $R=\frac{abc}{4\Delta }$.

Answer 1

$a\cos B\cos C+b\cos C\cos A+c\cos A\cos B$

$=\cos C(a\cos B+b\cos A)+c\cos A\cos B$

$=\cos C\left(c\right)+c\cos A\cos B$ ....... [Since $a\cos B+b\cos A=c$]

$=c(\cos C+\cos A\cos B)$

$=c(\cos \{\pi -(A+B\left)\right\}+\cos A\cos B)$ ...... [Since $A+B+C=\pi$]

$=c(-\cos \left\{\right(A+B\left)\right\}+\cos A\cos B)$

$=c\left(\sin A\sin B\right)$

$=c.\frac{a}{2R}.\frac{b}{2R}$ ...... [Since, $\sin A=\frac{a}{2R}$ and $\sin B=\frac{b}{2R}$]

$=\frac{abc}{4R^2}$

$=\frac{\frac{abc}{4R}}{R}$

$=\frac{\Delta }{R}$.