If $sin A=\dfrac{3}{5},\tan B=\dfrac{1}{2}$ and $\dfrac{\pi}{2}<A<\pi<B<\dfrac{3\pi}{2}$, then the value of $8\tan A-\sqrt{5}sec B=$

Given $\dfrac{\pi}{2}<A<\pi,\implies \tan A<0$

$\sin A=\dfrac{3}{5}\implies \tan A=-\dfrac{3}{\sqrt{5^2-3^2}}=-\dfrac{3}{4}$

$\pi<B<\dfrac{3\pi}{2}\implies \text{sec} B<0$

$\tan B=\dfrac{1}{2}\implies \text{sec} B=-\sqrt{1+\dfrac{1}{4}}=-\dfrac{\sqrt{5}}{2}$

So $8\tan A-\sqrt{5}\text{sec}B=8\times \bigg(-\dfrac{3}{4}\bigg)-\sqrt{5}\times \bigg(-\dfrac{\sqrt{5}}{2}\bigg)=-6+\dfrac{5}{2}=-\dfrac{7}{2}$