Question
With usual notations prove that $2\left\{a \sin ^2 \frac{C}{2}+c \sin ^2 \frac{A}{2}\right\}=a-b+c$.
$=a\left(2 \sin ^2 \frac{C}{2}\right)+c\left(2 \sin ^2 \frac{\mathrm{A}}{2}\right)$
$=a(1-\cos \mathrm{C})+c(1-\cos \mathrm{A})$
$=a\left[1-\frac{a^2+b^2-c^2}{2 a b}\right]+c\left[1-\frac{b^2+c^2-a^2}{2 b c}\right]$
... [By cosine rule]
$=a\left[\frac{2 a b-a^2-b^2+c^2}{2 a b}\right]+c\left[\frac{2 b c-b^2-c^2+a^2}{2 b c}\right]$
$=\frac{2 a b-a^2-b^2+c^2}{2 b}+\frac{2 b c-b^2-c^2+a^2}{2 b}$
$=\frac{2 a b-a^2-b^2+c^2+2 b c-b^2-c^2+a^2}{2 b}$
$=\frac{2 a b-2 b^2+2 b c}{2 b}$
$=a-b+c=$ RHS
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| $x$ | $0$ | $1$ | $2$ | $3$ | $4$ |
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$\cos ^{-1} x=\tan ^{-1} \frac{\sqrt{1-x^2}}{x}$, if $x<0$.
Question is modified
$\cos ^{-1} x=\tan ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$, if $x>0$.