Isn't there one $y$ instead of $\alpha$?

$ \beta=\frac{\pi sinx}{4sin\alpha},\gamma=\frac{\pi cosx}{4cos\alpha}\in{[0,\frac{\pi}{2})} $,and $ \beta+\gamma=\frac{\pi sin(x+\alpha)}{2sin2\alpha}>\frac{\pi}{4} $ so $ tan\beta+tan\gamma>tan\frac{\pi}{4}=1 $ (where we have $ sin(x+\alpha)>\frac{1}{2},sin2\alpha\leq 1 $)

how $\tan\beta+\tan\gamma>tan\frac{\pi}{4}=1$ does hold?

Pantum wrote: $ \beta=\frac{\pi sinx}{4sin\alpha},\gamma=\frac{\pi cosx}{4cos\alpha}\in{[0,\frac{\pi}{2})} $,and $ \beta+\gamma=\frac{\pi sin(x+\alpha)}{2sin2\alpha}>\frac{\pi}{4} $ so $ tan\beta+tan\gamma>tan\frac{\pi}{4}=1 $ (where we have $ sin(x+\alpha)>\frac{1}{2},sin2\alpha\leq 1 $) How to proof the ineq at line 2?

Since the required inequality holds for $x=\alpha$, therefore we just need to consider the case $x \ne \alpha$. Let $ A=\frac{\pi sinx}{4sin\alpha}$ and $B=\frac{\pi cosx}{4cos\alpha}$. Obviously $A,B\in{[0,\frac{\pi}{2})}$. If one of $A$ and $B$ is greater than $\frac{\pi}{4}$,then the desired inequality follows immediately. Now we shall prove that one of A and B is greater than $\frac{\pi}{4}$, which is equivalent to show $\frac{sinx}{sin\alpha}$ or $\frac{cosx}{cos\alpha}$ must be greater than 1. Suppose $\frac{sinx}{sin\alpha}< 1$ and $\frac{cosx}{cos\alpha}< 1$. From the first inequality we have $x< \alpha$, while the second one imples $x> \alpha$, which is a contradiction. Hence one of $A$ and $B$ is greater than $\frac{\pi}{4}$.