Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively, then, $\cos A+\cos B+\cos C$ is equal to

• A. $1+\frac{r}{R}$
• B. $1-\frac{r}{R}$
• C. $1+\frac{R}{r}$
• D. $1-\frac{R}{r}$

Answer 1

The correct option is A $1+\frac{r}{R}$

The value of $\cos A+\cos B+\cos C$

$=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}$

$=1+2\sin \frac{C}{2}[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)](\because \frac{\pi }{2}=\frac{A+B}{2}-\frac{C}{2})$

$=1+4\sin \frac{C}{2}\sin \frac{A}{2}\sin \frac{B}{2}(\because r=4R\sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2})$

$=1+\frac{r}{R}$