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

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
  1. $a^2$
  2. $c^2$
  3. $b^2$
  4. $a^2+b^2$

Solution

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} $

Asked in: AP EAMCET 2004

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