Question 49, Solution: Since $\cos A=\sin B=2 \tan \frac{C}{2}$, we have $B=90^{\circ} \pm A$, and $\mathrm{C}<90^{\circ}$. If $\mathrm{B}=90^{\circ}-\mathrm{A}$, then $\mathrm{C}=90^{\circ}$, which is a contradiction. Therefore, $\mathrm{B}=90^{\circ}+\mathrm{A}, \mathrm{C}=90^{\circ}-2 \mathrm{~A}$. Hence
$$
\begin{array}{l}
\frac{\cos A}{2}=\tan \frac{C}{2}=\tan \left(45^{\circ}-A\right)=\frac{1-\tan A}{1+\tan A}=\frac{\cos A-\sin A}{\cos A+\sin A} \\
\Rightarrow \cos ^{2} A+\sin A \cos A=2 \cos A-2 \sin A \\
\Rightarrow \sin A+\cos A+2 \tan A=2
\end{array}
$$

Therefore, the value of $\sin A+\cos A+2 \tan A$ is 2.
Question 49, if the three interior angles $A, B, C$ of $\triangle ABC$ satisfy $\cos A = \sin B = 2 \tan \frac{C}{2}$, then the value of $\sin A + \cos A + 2 \tan A$ is $\qquad$ -