Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

What is the value of $\left(2ac\right)\sin \frac{A-B+C}{2}$ is equal to?

• 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

Answer: (B)

$c^2+a^2-b^2$

we know that,

$A+B+C={180}^o$

$\frac{A+B+C}{2}={90}^o$

$\frac{A+C-B}{2}={90}^o-B\Rightarrow \sin \frac{A+C-B}{2}=\sin ({90}^o-B)=\cos B$

$\left(2ac\right)\sin \frac{A+C-B}{2}=\left(2ac\right)\cos B=2ac\frac{c^2+a^2-b^2}{2ac}=c^2+a^2-b^2$

Therefore, Answer is $B$