Question

In a triangle $ABC$ , the value of $(\cos A+\cos B+\cos C)$ is

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

Answer 1

Answer: (D)

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

Let $t=\cos A+\cos B+\cos C$
$\Rightarrow t=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}$
$\Rightarrow t=1+2\sin \frac{C}{2}\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}$
$\Rightarrow t=1+2\sin \frac{C}{2}(\cos \left(\frac{A-B}{2}\right)-\sin \frac{C}{2})=1+2\sin \frac{C}{2}(\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right))$
$\Rightarrow t=1+4\sin \frac{C}{2}\sin \frac{A}{2}\sin \frac{B}{2}=1+\frac{1}{R}\left(4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\right)$
$\Rightarrow t=1+\frac{r}{R}$
Ans: D