Question

In a $\Delta ABC$ the value of $\left(\frac{r_1+r_2}{1+\cos C}\right)$ is always equal to

• A. $2r$
• B. $2R$
• C. $\frac{2r^2}{R}$
• D. $\frac{2R^2}{r}$

Answer 1

Answer: (B)

$2R$

We know that, In a triangle ABC

$r_1=4R\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

$r_2=4R\cos \frac{A}{2}\sin \frac{B}{2}\cos \frac{C}{2}$

$\Rightarrow r_1+r_2=4R(\sin \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}+\sin \frac{B}{2}\cos \frac{A}{2}\cos \frac{C}{2})$

$\Rightarrow r_1+r_2=4R\cos \frac{C}{2}(\sin \frac{A}{2}\cos \frac{B}{2}+\sin \frac{B}{2}\cos \frac{A}{2})$

$\Rightarrow r_1+r_2=4R\cos \frac{C}{2}\sin \left(\frac{A+B}{2}\right)$ $[\because \sin (P+Q)=\sin P\cos Q+\cos P\sin Q]$

$\Rightarrow r_1+r_2=4R\cos \frac{C}{2}\sin \left(\frac{\pi -C}{2}\right)$ $[\because A+B+C=\pi ]$

$\Rightarrow r_1+r_2=4R\cos \frac{C}{2}\sin (\frac{\pi }{2}-\frac{C}{2})$

$\Rightarrow r_1+r_2=4R{\cos }^2\frac{C}{2}$

$\Rightarrow r_1+r_2=2R(1+\cos C)$ $[\because \cos 2\theta =2{\cos }^2\theta -1]$

$\Rightarrow \frac{r_1+r_2}{1+\cos C}=2R$