If (a+2) α+(2 a-1) α=2 a+1 , where 0 α (π)/(2) and a 1 , then α is...

$$\begin{aligned} & (a+2) \frac{2 \tan \frac{\alpha}{2}}{1+\tan ^2 \frac{\alpha}{2}}+(2 a-1) \frac{\left(1-\tan ^2 \frac{\alpha}{2}\right)}{\left(1+\tan ^2 \frac{\alpha}{2}\right)} \\ & =2 a+1 \end{aligned}$$ Let $\tan \frac{\alpha}{2}=\mathrm{t}$$\begin{aligned} & (\mathrm{a}+2) 2 \mathrm{t}+(2 \mathrm{a}-1)\left(1-\mathrm{t}^2\right)=(2 \mathrm{a}+1)\left(1+\mathrm{t}^2\right) \\ & \Rightarrow 4 \mathrm{a} \cdot \mathrm{t}^2-2 \mathrm{t}(\mathrm{a}+2)+2=0 \\ & \Rightarrow \mathrm{t}^2-\mathrm{t}\left(\frac{1}{2}+\frac{1}{\mathrm{a}}\right)+\frac{1}{2 \mathrm{a}}=0 \\ & \Rightarrow \quad\left(\mathrm{t}-\frac{1}{2}\right)\left(\mathrm{t}-\frac{1}{\mathrm{a}}\right)=0 \\ & \therefore \quad \tan \frac{\alpha}{2}=\frac{1}{2} \quad \text { or } \quad \tan \frac{\alpha}{2}=\frac{1}{\mathrm{a}} \end{aligned}$If $\tan \frac{\alpha}{2}=\frac{1}{2} \quad$ or If $\quad \tan \frac{\alpha}{2}=\frac{1}{a}\begin{aligned} & \therefore \quad \tan \alpha=\frac{2 \times \frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3} \\ & \Rightarrow \quad \tan \alpha=\frac{2 \times \frac{1}{a}}{1-\frac{1}{a^2}}=\frac{2 a}{a^2-1} .\end{aligned}$$$