Question

In $∆ABC,{(a-b)}^2{\cos }^2\frac{C}{2}+{(a+b)}^2{\sin }^2\frac{C}{2}$ is equal to

• A. $a^2$
• B. $b^2$
• C. $c^2$
• D. None of these

Answer 1

The correct option is C

$c^2$

Explanation for the correct option:

Solving the expression:

Given

$$\begin{gathered} {(a-b)}^2{\cos }^2\frac{C}{2}+{(a+b)}^2{\sin }^2\frac{C}{2}=\left(a^2+b^2-2ab\right){\cos }^2\frac{C}{2}+\left(a^2+b^2+2ab\right){\sin }^2\frac{C}{2} \\ =(a^2+b^2)\left({\cos }^2\frac{C}{2}+{\sin }^2\frac{C}{2}\right)-2ab\left({\cos }^2\frac{C}{2}-{\sin }^2\frac{C}{2}\right) \end{gathered}$$

Using the identity ${\sin }^{2}A+{\cos }^{2}A=1$.

$=(a^2+b^2)-2ab\left({\cos }^2\frac{C}{2}-{\sin }^2\frac{C}{2}\right)$

Now, ${\cos }^{2}A-{\sin }^{2}A=\cos 2A$

$$\begin{gathered} \Rightarrow {\cos }^2\frac{C}{2}-{\sin }^2\frac{C}{2}=\cos \left(2\frac{C}{2}\right) \\ \Rightarrow {\cos }^2\frac{C}{2}-{\sin }^2\frac{C}{2}=\cos C \end{gathered}$$

${\therefore (a-b)}^2{\cos }^2\frac{C}{2}+{(a+b)}^2{\sin }^2\frac{C}{2}=(a^2+b^2)-2ab\cos C$

We know in $∆ABC$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

Substituting the value of $\cos C$ in the above equation, we get

$$\begin{gathered} {\therefore (a-b)}^2{\cos }^2\frac{C}{2}+{(a+b)}^2{\sin }^2\frac{C}{2}=(a^2+b^2)-2ab\left({\cos }^2\frac{C}{2}-{\sin }^2\frac{C}{2}\right) \\ =(a^2+b^2)-2ab\cos C \\ =(a^2+b^2)-2ab\frac{\left(a^2+b^2-c^2\right)}{2ab} \\ =a^2+b^2-a^2-b^2+c^2 \\ =c^2 \end{gathered}$$

Hence, option (C) is the correct answer.