Question

Prove that acosbcosc + bcosccosa + ccosacosb = abc/4r2

Answer 1

Proved below.

Step-by-step explanation:

Given:

$acosbcosc + bcosccosa + ccosacosb = \frac{abc}{4r^2}$

LHS =  acosbcosc + bcosccosa + ccosacosb

      = $cosacosbcosc(\frac{acosbcosc}{cosacosbcosc} +\frac{bcosccosa}{cosacosbcosc} +\frac{ccosacosb}{cosacosbcosc})$    [Divide and multiply by cosacosbcosc]

        = cosacosbcosc[$\frac{a}{cosa}+ \frac{b}{cosb}+ \frac{c}{cosc}$]

 

Using Sine rule:

= cosacosbcosc(2r)[tana + tanb + tanc]      [R= circumradius ]

= cosacosbcosc(2r)[tanatanbtanc]       [tana+tanb+tanc = tanatanbtanc]

= $cosacosbcosc(2r)[\frac{sinasinbsinc}{cosacosbcosc} ]$

= 2r[sinasinbsinc]

Again using the sine rule, we get

= $\frac{2rabc}{2r\times2r\times2r}$

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

= RHS

Hence proved.