$\text{ If }\int \frac{dx}{{\sin }^{2}x+{\tan }^{2}x}=A\cot x+B{\tan }^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)+c\text{ then }\frac{A}{B}=$

$\int \frac{dx}{{\sin }^{2}x+{\tan }^{2}x}=\int \frac{{\sec }^{2}x}{{\tan }^{2}x+{\tan }^{2}x{\sec }^{2}x}dx$

$=\int \frac{{\sec }^{2}x}{{\tan }^{2}x+{\tan }^{2}x\left(1+{\tan }^{2}x\right)}$

$=\int \frac{dt}{t^4+2t^2} put \tan x =t \Rightarrow se{c}^{2}x dx =dt$

$\begin{array}{l}=\int \frac{dt}{t^2\left(t^2+2\right)}\\ =\frac{1}{2}\int \frac{\left(t^2+2\right)-t^2}{t^2\left(t^2+2\right)}dt\\ =\frac{1}{2}\int \left(\frac{1}{t^2}-\frac{1}{t^2+2}\right)dt\\ =\frac{1}{2}\left[\frac{-1}{t}-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{t}{\sqrt{2}}\right]+c\\ =-\frac{1}{2}\cot x-\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)tc\end{array}$

$A=-1/2,B=\frac{-1}{2\sqrt{2}}$

$\frac{A}{B}=\sqrt{2}$