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Let \(a, b\) and \(c\) be the lengths of the sides of a triangle with its opposite angles \(A, B\) and \(C\) respectively. If \(\angle C=60^{\circ}\), then the value of \(\frac{c(a+b)+\left(a^2+b^2\right)}{(b+c)(c+a)}\) is
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\(\begin{aligned}
& \cos C=\frac{a^2+b^2-c^2}{2 a b} \\
& \Rightarrow \quad \cos 60^{\circ}=\frac{a^2+b^2-c^2}{2 a b} \\
& \Rightarrow a^2+b^2-c^2=a b \\
& \Rightarrow \quad a^2+b^2=a b+c^2 \\
& \text {Now, } \frac{c(a+b)+\left(a^2+b^2\right)}{(b+c)(c+a)}=\frac{c a+b c+a b+c^2}{b c+a b+c^2+a c}=1
\end{aligned}\)
& \cos C=\frac{a^2+b^2-c^2}{2 a b} \\
& \Rightarrow \quad \cos 60^{\circ}=\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 \\
& \text {Now, } \frac{c(a+b)+\left(a^2+b^2\right)}{(b+c)(c+a)}=\frac{c a+b c+a b+c^2}{b c+a b+c^2+a c}=1
\end{aligned}\)
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