$\begin{array}{l}A:{\sin }^{2}5°+{\sin }^{2}10°+\mathrm{.....}+{\sin }^{2}85°+{\sin }^{2}90°\\ =({\sin }^{2}5°+{\sin }^{2}85°)+({\sin }^{2}10°+{\sin }^{2}80°)+({\sin }^{2}15°+{\sin }^{2}75°)\\ +({\sin }^{2}20°+{\sin }^{2}70°)+({\sin }^{2}25°+{\sin }^{2}65°)+({\sin }^{2}30°+{\sin }^{2}60°)\\ +({\sin }^{2}35°+{\sin }^{2}55°)+({\sin }^{2}40°+{\sin }^{2}50°)+({\sin }^{2}45°+{\sin }^{2}90°)\end{array}$

But we know that  $\sin (90°-\theta )=\cos \theta$

$\sin 85°=\sin (90°-5°)=\cos 5°$

Similarly $\sin 80°=\cos 10°$

$\sin 75°=\cos 15°\mathrm{......}$

$\begin{array}{l}=({\sin }^{2}5°+{\cos }^{2}5°)+({\sin }^{2}10°+{\cos }^{2}10°)+({\sin }^{2}15°+{\cos }^{2}15°)\\ +({\sin }^{2}20°+{\cos }^{2}20°)+({\sin }^{2}25°+{\cos }^{2}25°)+({\sin }^{2}30°+{\cos }^{2}30°)\\ +({\sin }^{2}35°+{\cos }^{2}35°)+({\sin }^{2}40°+{\cos }^{2}40°)+({\sin }^{2}45°+{\sin }^{2}90°)\\ =1+1+1+1+1+1+1+1+{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(1\right)}^2\\ =9+\frac{1}{2}\\ =9+0.5=9.5\end{array}$

Statement B ${\sec }^2\theta +{\tan }^2\theta =1$

$(\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1$

$\frac{1}{p}(\sec \theta -\tan \theta )=1$

$\therefore \sec \theta -\tan \theta =p$