The proof of \$PF_{p}=cos\theta_p=sin\theta_s\$
In series RC, current passes through both R and C.

$$
P=i^2\cdot R \\ Q=i^2\cdot X_C \\ S = \sqrt{P^2+Q^2}=i^2\sqrt{R^2+X_C^2} \\ \therefore PF_s=\frac{P}{S}=\frac {R}{\sqrt{R^2+X_C^2}}
$$

In parallel RC (with the same R and C), the supply voltage is applied to both R and C.

$$
P = \frac{V^2}{R} \\ Q=\frac{V^2}{X_C} \\ S = \sqrt{P^2+Q^2}=V^2\sqrt{\frac{1}{R^2}+\frac{1}{X_C^2}}=V^2\cdot \frac{\sqrt{R^2+X_C^2}}{R \cdot X_C} \\ \therefore PF_p=\frac{P}{S}=\frac {X_C}{\sqrt{R^2+X_C^2}}
$$

So, finally we obtain 
$$
PF_s^2 + PF_p^2 = 1
$$