1. ## Equality !!

Hi !!

$\alpha$ , $\beta$ and $\gamma$ the angles of a triangle , prove that :

2. $\cos\alpha+\cos\beta+1-\cos\gamma=2\cos\frac{\alpha+\beta}{2}\cos\frac{\a lpha-\beta}{2}+2\sin^2\frac{\gamma}{2}=$

$=2\sin\frac{\gamma}{2}\cos\frac{\alpha-\beta}{2}+2\sin^2\frac{\gamma}{2}=2\sin\frac{\gamm a}{2}\left(\cos\frac{\alpha-\beta}{2}+\sin\frac{\gamma}{2}\right)=$

$=2\sin\frac{\gamma}{2}\left(\cos\frac{\alpha-\beta}{2}+\cos\frac{\alpha+\beta}{2}\right)=4\sin\ frac{\gamma}{2}\cos\frac{\alpha}{2}\cos\frac{\beta }{2}$

Then $\frac{\cos\alpha+\cos\beta}{\cos\alpha+\cos\beta+1-\cos\gamma}=\frac{2\sin\frac{\gamma}{2}\cos\frac{\ alpha-\beta}{2}}{4\sin\frac{\gamma}{2}\cos\frac{\alpha}{ 2}\cos\frac{\beta}{2}}=\frac{\cos\frac{\alpha-\beta}{2}}{2\cos\frac{\alpha}{2}\cos\frac{\beta}{2 }}$

$\frac{\cos\gamma+\cos\beta}{\cos\gamma+\cos\beta+1-\cos\alpha}=\frac{2\sin\frac{\alpha}{2}\cos\frac{\ gamma-\beta}{2}}{4\sin\frac{\alpha}{2}\cos\frac{\gamma}{ 2}\cos\frac{\beta}{2}}=\frac{\cos\frac{\gamma-\beta}{2}}{2\cos\frac{\gamma}{2}\cos\frac{\beta}{2 }}$

$\frac{\cos\alpha+\cos\gamma}{\cos\alpha+\cos\gamma +1-\cos\beta}=\frac{2\sin\frac{\beta}{2}\cos\frac{\al pha-\gamma}{2}}{4\sin\frac{\beta}{2}\cos\frac{\alpha}{ 2}\cos\frac{\gamma}{2}}=\frac{\cos\frac{\alpha-\gamma}{2}}{2\cos\frac{\alpha}{2}\cos\frac{\gamma} {2}}$

Then the left side member of the equality is

$\frac{\cos\frac{\alpha-\beta}{2}}{2\cos\frac{\alpha}{2}\cos\frac{\beta}{2 }}+\frac{\cos\frac{\beta-\gamma}{2}}{2\cos\frac{\beta}{2}\cos\frac{\gamma}{ 2}}+\frac{\cos\frac{\gamma-\alpha}{2}}{2\cos\frac{\gamma}{2}\cos\frac{\alpha} {2}}=$

$=\frac{\cos\frac{\gamma}{2}\cos\frac{\alpha-\beta}{2}+\cos\frac{\alpha}{2}\cos\frac{\beta-\gamma}{2}+\cos\frac{\beta}{2}\cos\frac{\gamma-\alpha}{2}}{2\cos\frac{\alpha}{2}\cos\frac{\beta}{ 2}\cos\frac{\gamma}{2}}=$

$=\frac{\cos\frac{\gamma+\alpha-\beta}{2}+\cos\frac{\gamma-\alpha+\beta}{2}+\cos\frac{\alpha+\beta-\gamma}{2}+\cos\frac{\alpha-\beta+\gamma}{2}+\cos\frac{\beta+\gamma-\alpha}{2}+\cos\frac{\beta-\gamma+\alpha}{2}}{4\cos\frac{\alpha}{2}\cos\frac{ \beta}{2}\cos\frac{\gamma}{2}}=$

$=\frac{2\cos\left(\frac{\pi}{2}-\beta\right)+2\cos\left(\frac{\pi}{2}-\alpha\right)+2\cos\left(\frac{\pi}{2}-\gamma\right)}{4\cos\frac{\alpha}{2}\cos\frac{\bet a}{2}\cos\frac{\gamma}{2}}=$

$=\frac{\sin\alpha+\sin\beta+\sin\gamma}{2\cos\frac {\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2} }=\frac{4\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\c os\frac{\gamma}{2}}{2\cos\frac{\alpha}{2}\cos\frac {\beta}{2}\cos\frac{\gamma}{2}}=2$