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In a $\triangle A B C,(a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2}$ is equal to
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Verified Answer
The correct answer is:
$c^2$
We have,
$$
\begin{aligned}
& (a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2} \\
& =\left(a^2+b^2-2 a b\right) \cos ^2 \frac{C}{2}+\left(a^2+b^2+2 a b\right) \sin ^2 \frac{C}{2} \\
& =a^2(1)+b^2(1)-2 a b\left(\cos ^2 \frac{C}{2}-\sin ^2 \frac{C}{2}\right) \\
& =a^2+b^2-2 a b \cos C \\
& {\left[\begin{array}{c}
\because \cos C=\frac{a^2+b^2-c^2}{2 a b} \text { or } \\
a^2+b^2-2 a b \cos C=c^2
\end{array}\right]} \\
& =C^2 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& (a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2} \\
& =\left(a^2+b^2-2 a b\right) \cos ^2 \frac{C}{2}+\left(a^2+b^2+2 a b\right) \sin ^2 \frac{C}{2} \\
& =a^2(1)+b^2(1)-2 a b\left(\cos ^2 \frac{C}{2}-\sin ^2 \frac{C}{2}\right) \\
& =a^2+b^2-2 a b \cos C \\
& {\left[\begin{array}{c}
\because \cos C=\frac{a^2+b^2-c^2}{2 a b} \text { or } \\
a^2+b^2-2 a b \cos C=c^2
\end{array}\right]} \\
& =C^2 \\
&
\end{aligned}
$$
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