Вычислить Tg\alpha/2 если sin\alpha + cos\alpha=\sqrt{7}/2

Решите задачу:

$sina+cosa=\frac{\sqrt7}{2}\\\\\frac{2tg\frac{a}{2}}{1+tg^2\frac{a}{2}}+\frac{1-tg^2\frac{a}{2}}{1+tg^2\frac{a}{2}}=\frac{\sqrt7}{2}\\\\t=tg\frac{a}{2}\; ,\; \; \frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2}-\frac{\sqrt7}{2}=0\; \; ,\; \; \frac{4t+2-2t^2-\sqrt7-\sqrt7t^2}{2(1+t^2)}=0\; ,\\\\(-2-\sqrt7)t^2+4t+(2-\sqrt7)=0\; ,\\\\(2+\sqrt7)t^2-4t-(2-\sqrt7)=0\\\\D/4=2^2+(2+\sqrt7)(2-\sqrt7)=4+4-7=1\\\\t_{1,2}=\frac{2\pm 1}{2+\sqrt7}\\\\tg\frac{a}{2}=\frac{2-1}{2+\sqrt7}=\frac{2-\sqrt7}{(2+\sqrt7)(2-\sqrt7)}=\frac{2-\sqrt7}{4-7}=\frac{2-\sqrt7}{-3}=\frac{\sqrt7-2}{3}$

$tg\frac{a}{2}=\frac{2+1}{2+\sqrt7}=\frac{3(2-\sqrt7)}{(2+\sqrt7)(2-\sqrt7)}=\frac{3(2-\sqrt7)}{4-7}=\frac{3(2-\sqrt7)}{-3}=\sqrt7-2$