$\dfrac{\pi}{2}<x<\pi\to \sin x>0,\cos x<0$
$\to \cos x=-\sqrt{1-\sin^2x}=\dfrac{-\sqrt{15}}{4}$
$\to \tan x=\dfrac{\sin x}{\cos x}=\dfrac{-1}{\sqrt{15}}$
$\to \cot x=\dfrac{1}{\tan x}=-\sqrt{15}$
$P=\cos(4\pi+\pi-x)+\tan(\pi+\dfrac{\pi}{2}+x\Big)$
$=\cos(\pi-x)+\tan\Big(\dfrac{\pi}{2}+x\Big)$
$=-\cos x-\cot x$
$=\dfrac{\sqrt{15}}{4}+\sqrt{15}=\dfrac{5\sqrt{15}}{4}$
