Q147 / 200 · TS EAMCET 2001 · Mathematics
In a $\triangle A B C, \frac{\cos C+\cos A}{c+a}+\frac{\cos B}{b}$ is equal to
- A$\frac{1}{a}$
- Correct answerB$\frac{1}{b}$
- C$\frac{c+a}{b}$
- D$1$
Solution
We have,
$\begin{aligned} & \frac{\cos C+\cos A}{c+a}=\frac{\cos C+\cos A}{2 R(\sin C+\sin A)} \\ = & \frac{2 \cos \frac{C+A}{2} \cos \frac{C-A}{2}}{2 R \cdot 2 \sin \frac{C+A}{2} \cos \frac{C-A}{2}}=\frac{1}{2 R} \tan \frac{B}{2}\end{aligned}$
Also,
$\frac{\cos B}{b}=\frac{\cos B}{2 R \sin B}=\frac{1}{2 R} \cdot \frac{1-\tan ^2 \frac{B}{2}}{2 \tan \frac{B}{2}}$
$\begin{aligned} & \therefore \quad \frac{\cos C+\cos A}{c+a}+\frac{\cos B}{b} \\ & =\frac{1}{2 R} \tan \frac{B}{2}+\frac{1}{2 R} \frac{1-\tan ^2 \frac{B}{2}}{2 \tan \frac{B}{2}} \\ & =\frac{1}{2 R}\left[\frac{2 \tan ^2 \frac{B}{2}+1-\tan ^2 \frac{B}{2}}{2 \tan \frac{B}{2}}\right]=\frac{1}{2 R}\left[\frac{1+\tan ^2 \frac{B}{2}}{2 \tan \frac{B}{2}}\right] \\ & =\frac{1}{2 R \sin B}=\frac{1}{b}\end{aligned}$
Hence, the correct answer is option: 'B'.