Paso 1: Sea $s=\sen^{2}x$ y $c=\cos^{2}x$, entonces $s+c=1$.
Además:
$$ \csc^{4}x=\frac{1}{s^{2}},\quad \sec^{4}x=\frac{1}{c^{2}},\quad \tan^{4}x=\frac{s^{2}}{c^{2}},\quad \cot^{4}x=\frac{c^{2}}{s^{2}}. $$

La condición dada queda:
$$ \frac{1}{c^{2}}+\frac{1}{s^{2}}=2K+\frac{s^{2}}{c^{2}}+\frac{c^{2}}{s^{2}}. $$
Reordenando:
$$ \frac{1-s^{2}}{c^{2}}+\frac{1-c^{2}}{s^{2}}=2K. $$
Como $1-s^{2}=(1-s)(1+s)=c(1+s)$ y $1-c^{2}=s(1+c)$:
$$ \frac{c(1+s)}{c^{2}}+\frac{s(1+c)}{s^{2}}=2K \Rightarrow \frac{1+s}{c}+\frac{1+c}{s}=2K. $$
Unificando:
$$ 2K=\frac{s(1+s)+c(1+c)}{sc} =\frac{(s+c)+(s^{2}+c^{2})}{sc} =\frac{1+(1-2sc)}{sc} =\frac{2-2sc}{sc} =2\frac{1-sc}{sc}. $$
Por tanto:
$$ K=\frac{1-sc}{sc}=\frac{1}{sc}-1 \Rightarrow \frac{1}{sc}=K+1 \Rightarrow sc=\frac{1}{K+1}. $$

Paso 2: Calcular $A$:
$$ A=\frac{1+s}{\csc^{4}x}+\frac{1+c}{\sec^{4}x} =(1+s)s^{2}+(1+c)c^{2}=s^{2}+s^{3}+c^{2}+c^{3}. $$
Usando:
$$ s^{2}+c^{2}=(s+c)^{2}-2sc=1-2sc,\qquad s^{3}+c^{3}=(s+c)^{3}-3sc(s+c)=1-3sc, $$
se tiene:
$$ A=(1-2sc)+(1-3sc)=2-5sc. $$
Sustituyendo $sc=\frac{1}{K+1}$:
$$ A=2-\frac{5}{K+1}=\frac{2(K+1)-5}{K+1}=\frac{2K-3}{K+1}. $$

Resultado final: $\boxed{A=\frac{2K-3}{K+1}}$.