Question

With the usual notations prove that
$2\{a{\sin }^2\frac{C}{2}+c{\sin }^2\frac{A}{2}\}=(a+c-b)$

Answer 1

Actually it is a wrong question. we can't prove this. It should be like this-
$2\{asi{n}^2\frac{C}{2}+c{\sin }^2\frac{A}{2}\}=(a+c-b)$
so,now we will prove this-
as we know-
${\sin }^2\theta =\frac{1-\cos 2\theta }{2}$
now we will apply this formula in our question-
L.H.S.
=$2\{asi{n}^2\frac{C}{2}+c{\sin }^2\frac{A}{2}\}$
=$2\{a\left(\frac{1-\cos 2\times \frac{C}{2}}{2}\right)+c\left(\frac{1-\cos 2\times \frac{A}{2}}{2}\right)\}$
=$a(1-\cos C)+c(1-\cos A)$
=$a+c-(a\cos C+c\cos A)$
here we will apply cosine rule$\Rightarrow (a\cos C+c\cos A)=b$
=$a+c-b$
=R.H.S.