Question

Mark the correct alternative in each of the following:

In any ∆ABC, the value of $2ac\sin \left(\frac{A-B+C}{2}\right)$ is

(a) $a^2+b^2-c^2$ (b) $c^2+a^2-b^2$ (c) $b^2-c^2-a^2$ (d) $c^2-a^2-b^2$

Answer 1

In ∆ABC,

$$\begin{gathered} A+B+C=\mathrm{\pi }\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right) \\ \Rightarrow A+C=\mathrm{\pi }-B \end{gathered}$$

$$\begin{gathered} \therefore 2ac\sin \left(\frac{A-B+C}{2}\right) \\ =2ac\sin \left(\frac{\mathrm{\pi }-2B}{2}\right) \\ =2ac\sin \left(\frac{\mathrm{\pi }}{2}-B\right) \\ =2ac\cos B \end{gathered}$$
$$\begin{gathered} =2ac\left(\frac{c^2+a^2-b^2}{2ca}\right)\left(\mathrm{Using}\mathrm{cosine}\mathrm{rule}\right) \\ =c^2+a^2-b^2 \end{gathered}$$

Hence, the correct answer is option (b).