Question
In a triangle $$ABC,$$ $$2ac\sin \frac{1}{2}\left( {A - B + C} \right) = $$
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 :
$${c^2} + {a^2} - {b^2}$$
Solution :
We know that $$A + B + C = 180°$$
⇒ $$A + C- B = 180° - 2B$$
Now $$2ac\sin \left[ {\frac{1}{2}\left( {A - B + C} \right)} \right]$$
$$\eqalign{
& = 2ac\sin \left( {{{90}^ \circ } - B} \right) \cr
& = 2ac\cos B \cr
& = \frac{{2ac\left( {{a^2} + {c^2} - {b^2}} \right)}}{{2ac}} \cr
& = {a^2} + {c^2} - {b^2} \cr} $$