Question
In $\triangle ABC$ with usual notations, prove that : $2\left\{a \sin ^2\left(\frac{ C }{2}\right)+c \sin ^2\left(\frac{ A }{2}\right)\right\}=(a+c-b)$
✓
Answer
$\sin ^2 \theta=\frac{1-\cos 2 \theta}{2}$
L.H.S $=2\left\{a \sin ^2\left(\frac{C}{2}\right)+c \sin ^2\left(\frac{A}{2}\right)\right\}$
$=2\left\{\frac{a(1-\cos C)}{2}+\frac{c(1-\cos A)}{2}\right\}$
=a-acosC+c-ccosA
=(a+c)-(acosC+ccosA)
=a+c-b
R.H.S
$2 a \left\{\sin ^2\left(\frac{C}{2}\right)+c \sin ^2\left(\frac{A}{2}\right)\right\}=( a + c - b )$
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