A, B, C

$\begin{aligned} \text{Since } \quad A + B + C & = 180^\circ \\ A + B + 90^ \circ & = 180^\circ \\ \Rightarrow B & = 90^\circ - A \end{aligned}$

$\begin{aligned} \text{It is given that } \quad 2\tan A & = \sin B \\ 2\frac{\sin A}{\cos A} & = \sin (90^\circ - A) \\ 2\frac{\sin A}{\cos A} & = \cos A \\ 2\sin A & = \cos^2 A \\ 2\sin A & = 1 - \sin^2 A \\ \Rightarrow \sin^2 A + 2 \sin A - 1 & = 0 \\ \sin A & = \frac{-2\pm \sqrt{4+4}}{2}\\ & = - 1 + \sqrt{2} \quad \quad \quad \quad \quad \quad \small \color{#D61F06}{|-1-\sqrt{2}| > 1 \text{ is rejected.}}\\ & = 2 \left(\frac{1}{\sqrt{2}}\right) - 1 \\ & = \boxed{2\cos 45^\circ - \sin 90^\circ} \end{aligned}$

Since ACB = 90, it implies that A + B = 90. Hence, sin B = cos A So, the given equation may be rewritten as 2 tan A = cos A. Multiply both sides by cos A. We, therefore, have 2 sin A = (cos A)^2 Let sin A = x. Therefore, the above equation may be further rewritten as 2x = 1-x^2 Solving this would yield x = root(2) - 1. (Since sine is range bound to [-1,+1], the other root of the quadratic - root(2) - 1 is ruled out). 2 cos 45 - sin 90 = 2*1/root(2) - 1 = root(2) - 1