If  $f\left(x\right)=\left\{\begin{array}{l}\begin{array}{ccc}\frac{1-{\sin }^{3}x}{3{\cos }^{2}x}& if& x<\frac{\pi }{2}\end{array}\\ \begin{array}{c}\begin{array}{ccc}p& if& x=\frac{\pi }{2}\end{array}\\ \\ \begin{array}{ccc}\frac{4q(1-\sin x)}{{(\pi -2x)}^2}& if& x>\frac{\pi }{2}\end{array}\end{array}\end{array}\right.$  is continuous at $x=\frac{\pi }{2},$  then (p,q) =

$$\begin{gathered} LHL=\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }\frac{1-{\sin }^{3}x}{3{\cos }^{2}x}=\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\lim }\frac{\left(1-\sin x\right)\left(1+\sin x+{\sin }^{2}x\right)}{3(1-\sin x)(1+\sin x)}=\frac{3}{3\left(2\right)}=\frac{1}{2} \\ RHL=\underset{X\to {\frac{\mathrm{\pi }}{2}}^{+}}{\lim }\frac{4q(1-\sin x)}{{(\mathrm{\pi }-2\mathrm{x})}^2}=\underset{X\to {\frac{\mathrm{\pi }}{2}}^{+}}{\lim }\frac{4q(1-\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}{4{\left(\frac{\mathrm{\pi }}{2}-x\right)}^2}=q\frac{1}{2}=\frac{q}{2} \\ LHL=f\left(\frac{\mathrm{\pi }}{2}\right)=RHL \\ \frac{1}{2}=P=\frac{q}{2}\Rightarrow p=\frac{1}{2,}q=1 \end{gathered}$$