Q66 / 160 · AP EAMCET 2024 (18 May Shift 1) · Mathematics

If $\triangle A B C, \cos A+\cos B+\cos C=$

  • A
    $1+\frac{r}{2 R}$
  • B
    $1-\frac{r}{\mathrm{R}}$
  • C
    $1+\frac{\mathrm{R}}{r}$
  • Correct answer
    D
    $1+\frac{r}{\mathrm{R}}$

Solution

Since,$\cos A+\cos B+\cos C$ $\begin{aligned} & =2 \sin \left(\frac{\mathrm{C}}{2}\right) \cdot \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)+1-2 \sin ^2\left(\frac{\mathrm{C}}{2}\right) \\ & =1+2 \sin \left(\frac{\mathrm{C}}{2}\right)\left(\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)-\sin \left(\frac{\mathrm{C}}{2}\right)\right)\end{aligned}$ $\begin{aligned} & =1+2 \sin \left(\frac{\mathrm{C}}{2}\right)\left(\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)-\cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\right) \\ & =1+2 \sin \left(\frac{\mathrm{C}}{2}\right)\left\{2 \sin \left(\frac{\mathrm{~A}}{2}\right) \sin \left(\frac{\mathrm{B}}{2}\right)\right\}\end{aligned}$ $\begin{aligned} & =1+4 \sin \left(\frac{\mathrm{~A}}{2}\right) \sin \left(\frac{\mathrm{B}}{2}\right) \cdot \sin \left(\frac{\mathrm{C}}{2}\right) \\ & =1+\frac{r}{\mathrm{R}} \quad\left\{\text { Since, } r=4 \mathrm{R} \sin \left(\frac{\mathrm{A}}{2}\right) \sin \left(\frac{\mathrm{B}}{2}\right) \sin \left(\frac{\mathrm{C}}{2}\right)\right\}\end{aligned}$

Hence, the correct answer is option: 'D'.

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