Question
In $\triangle ABC$ with the usual notations prove that $(a-b)^2 \cos ^2\left(\frac{ C }{2}\right)+(a+b)^2 \sin ^2\left(\frac{ C }{2}\right)=c^2$
$LHS =(a-b)^2 \cos ^2\left(\frac{C}{2}\right)+(a+b)^2 \sin ^2\left(\frac{C}{2}\right)$
$=a^2\left[\cos ^2\left(\frac{C}{2}\right)+\sin ^2\left(\frac{C}{2}\right)\right]+b^2\left[\cos ^2\left(\frac{C}{2}\right)+\sin ^2\left(\frac{C}{2}\right)\right]-2 a b\left[\cos ^2\left(\frac{C}{2}\right)-\sin ^2\left(\frac{C}{2}\right)\right]$
$=a^2+b^2-a^2-b^2+c^2$
$=c^2$
=RHS
Hence proved
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| $X =x$ | $1$ | $2$ | $3$ |
| $P ( X =x)$ | $\frac{1}{5}$ | $\frac{2}{5}$ | $\frac{2}{5}$ |
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