Let's determine the minimum of the function defined in the interval $45^{\circ}<x<90^{\circ}$.

$$
y=\operatorname{tg} x+\frac{\operatorname{tg} x}{\sin \left(2 x-90^{\circ}\right)}
$$
With identical transformations,

\[
\begin{aligned}
y= & \operatorname{tg} x\left(1-\frac{1}{\cos 2 x}\right)=\operatorname{tg} x \frac{\left(1-2 \sin ^{2} x\right)-1}{\cos ^{2} x-\sin ^{2} x}=\frac{2 \sin ^{3} x}{\cos x\left(\sin ^{2} x-\cos ^{2} x\right)}= \\
& \frac{2}{\operatorname{ctg} x-\operatorname{ctg}^{3} x},
\end{aligned}
\]

and since in the prescribed interval, according to the original form, \( y > 2 \operatorname{tg} x > 2 \), instead of the original question, we can seek

\[
\frac{2}{y}=\operatorname{ctg} x-\operatorname{ctg}^{3} x=\operatorname{ctg} x\left(1-\operatorname{ctg}^{2} x\right) \quad (0 < \operatorname{ctg} x < 1),
\]

where the function \( f(u) = u - u^3 \) (with \( u = \operatorname{ctg} x \)) has a maximum at \( u = \frac{1}{\sqrt{3}} \), and the minimum of \( y = \frac{2}{f(u)} \) is \( y = 3 \).

1. We can also reach the goal without squaring, if we factorize \( \frac{2}{y} \) into linear factors, then multiply these by positive constants, which we choose so that the sum of the resulting factors is independent of \( u \) and there is a value of \( u \) for which all three factors are equal. The first condition is satisfied by choosing factors of the form \( 1, \alpha, 1+\alpha \):

\[
\begin{aligned}
& \frac{2 \alpha(1+\alpha)}{y}=u \cdot[\alpha(1+u)] \cdot[(1+\alpha)(1-u)] \leqq \\
\leqq & \left(\frac{u+\alpha(1+u)+(1+\alpha)(1-u)}{3}\right)^{3}=\frac{(1+2 \alpha)^{3}}{27}
\end{aligned}
\]

Moreover, the system of equations

\[
\begin{aligned}
& u=\alpha(1+u) \\
& u=(1+\alpha)(1-u)
\end{aligned}
\]

must be solvable. From this, eliminating \( u \), we get the equation \( 2 \alpha^{2}+2 \alpha-1=0 \), whose root is \( \alpha=\frac{\sqrt{3}-1}{2} \), and with this, \( u=\frac{1}{\sqrt{3}} \), as in the above solution.