MHB What Is the Minimum Value of (a²+b²)/c² in Triangle ABC?

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The problem involves finding the minimum value of the expression (a²+b²)/c² in triangle ABC, given the equation involving the angles A and B. The equation states that the cosine of angle A divided by one plus the sine of angle A equals the sine of double angle B divided by one plus the cosine of double angle B. Participants are encouraged to explore relationships between the sides and angles of the triangle to derive the minimum value. The discussion emphasizes the importance of accurate calculations and logical reasoning in solving the problem. The goal is to determine this minimum value effectively within the constraints of triangle geometry.
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Here is this week's POTW:

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In a triangle $ABC$, it is given that $\dfrac{\cos A}{1+\sin A}=\dfrac{\sin 2B}{1+\cos 2B}$.

Find the minimum value of $\dfrac{a^2+b^2}{c^2}$.

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I hope I made no errors in calculation.
From laws of sines
f:=\frac{a^2+b^2}{c^2}=\frac{\sin^2 A+\sin^2 B}{\sin^2 C}=\frac{\sin^2 A+\sin^2 B}{\sin^2 (A+B)}
The given condition reads
\frac{\cos A}{1+\sin A}=\tan B
By this we can delete B in the above formula to get
f(x)=2x-3+\frac{4}{1+x}
where
x=\sin A

f'(x)=2-\frac{4}{(1+x)^2}
f'(\sqrt{2}-1)=0
f(\sqrt{2}-1)=4\sqrt{2}-5 \approx 0.66 as minimum.