Q65 / 160 · TS EAMCET 2015 · Mathematics
In $\triangle A B C$, if $\angle C=\frac{\pi}{3}$, then $\frac{3}{a+b+c}-\frac{1}{a+c}$ equals
- A$\frac{1}{a+b}$
- Correct answerB$\frac{1}{b+c}$
- C$\frac{1}{2 a+b}$
- D$\frac{1}{b+2 c}$
Solution
We have, $\angle C=\frac{\pi}{3}$
Now, $\quad \cos C=\frac{a^2+b^2-c^2}{2 a b}$
$$
\begin{aligned}
& \Rightarrow \quad \cos \frac{\pi}{3}=\frac{a^2+b^2-c^2}{2 a b} \\
& \Rightarrow a^2+b^2-c^2=a b \\
& \Rightarrow \quad a^2+b^2=a b+c^2 \\
& \Rightarrow a^2+b^2+a c+b c=a b+c^2+a c+b c
\end{aligned}
$$
[adding both sides by $a c+b c$ ]
$$
\begin{aligned}
& \Rightarrow \quad a(c+a)+b(b+c)=(b+c)(c+a) \\
& \Rightarrow \quad \frac{a}{b+c}+\frac{b}{c+a}=1 \\
& \Rightarrow \quad \frac{a}{b+c}+1+\frac{b}{c+a}+1=3 \\
& \Rightarrow \quad \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=3
\end{aligned}
$$
[adding 2 on both sides]
$\begin{array}{ll}\Rightarrow & \frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c} \\ \Rightarrow & \frac{3}{a+b+c}-\frac{1}{a+c}=\frac{1}{b+c}\end{array}$
Hence, the correct answer is option: 'B'.