If $\frac{2 x+1}{(x-1)^2\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C x+D}{x^2+1}$, then $A+B+C+D=$

Let $a$ be maximum value of $(3 \cos \theta-4 \sin \theta)$ and $\theta \neq \frac{n \pi}{2}$. If $\alpha=a \sin ^2 \theta \cdot \cos ^3 \theta$ and $\beta=a \sin ^3 \theta \cdot \cos ^2 \theta$, then $\sqrt{\frac{\left(\alpha^2+\beta^2\right)^5}{(\alpha \beta)^4}}=$

If $A$ does not belong to the first quadrant, $B$ does not belong to the second quadrant, $\sin A=\frac{11}{61}$ and $\cos B=\frac{-7}{25}$, then $A-B$ and $A+B$ lie respectively in the quadrants

If $\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x =\cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x$, then a possible value of $\sec x$ is