If $\frac{4 x^2+5 x^4+7}{\left(x^2+1\right)\left(x^4+x^2+1\right)}=\frac{A x+B}{x^2+1} +\frac{C x^3+D x^2+E x+F}{x^4+x^2+1}$, then $B+2(D+F+E)-C \cdot A=$
$$ \begin{aligned} \sin ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}-\sin ^4 \frac{3 \pi}{8} & +\sin ^4 \frac{5 \pi}{8} +\cos ^4 \frac{7 \pi}{8}-\sin ^4 \frac{7 \pi}{8}= \end{aligned} $$
$$ \operatorname{cosec}^{-1}\left[\left(\frac{\tan ^2\left(\frac{\alpha-\pi}{4}\right)-1}{\tan ^2\left(\frac{\alpha-\pi}{4}\right)+1}+\cos \frac{\alpha}{2} \cdot \cot 5 \alpha\right) \sec \frac{11 \alpha}{2}\right] $$
Assertion (A) If $A=15^{\circ}, B=17^{\circ}$ and $C=13^{\circ}$, then $\cot 2 A+\cot 2 B+\cot 2 C=\cot 2 A \cot 2 B \cot 2 C$
Reason (R) In a $\triangle P Q R$,
$$ \tan \frac{P}{2} \tan \frac{Q}{2}+\tan \frac{Q}{2} \tan \frac{R}{2}+\tan \frac{P}{2} \tan \frac{R}{2}=1 $$
The correct option among the following is
TS EAMCET Papers
All year-wise previous year question papers