MCQ
In $\triangle A B C$, with usual notations, $2 a c \sin \left(\frac{1}{2}(A-B+C)\right)$ is equal to
  • A
    $a^2+b^2-c^2$
  • $c^2+a^2-b^2$
  • C
    $b^2-c^2-a^2$
  • D
    $c^2-a^2-b^2$

Answer

Correct option: B.
$c^2+a^2-b^2$
(b) : In $\triangle A B C$, we have $A+B+C=180^{\circ}$ ...(i)
Now, $2 a c \sin \left(\frac{A-B+C}{2}\right)$
$
\begin{aligned}
& =2 a c \sin \left(\frac{180^{\circ}-2 B}{2}\right) \quad \text { [Using (i)] } \\
& =2 a c \sin \left(90^{\circ}-B\right)=2 a c \cos B \\
& =2 a c\left(\frac{c^2+a^2-b^2}{2 a c}\right)=c^2+a^2-b^2
\end{aligned}
$

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