Let $f:(0, \pi) \rightarrow \mathbf{R}$ be a function given by $$f(x)=\left\{\begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0< x<\frac{\pi}{2} \\ \mathrm{a}-8, & x=\frac{\pi}{2} \\ (1+\mid \cot x)^{\frac{\mathrm{b}}{\mathrm{a}}|\tan x|}, & \frac{\pi}{2} < x < \pi\end{array}\right.$$

where $\mathrm{a}, \mathrm{b} \in \mathbf{Z}$. If $f$ is continuous at $x=\frac{\pi}{2}$, then $\mathrm{a}^2+\mathrm{b}^2$ is equal to _________.

<p>$$\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} f(x)=f\left(\frac{\pi}{2}\right)=\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}} f(x) \text { for continuity at } x=\frac{\pi}{2}$$</p> <p>$$\begin{aligned} & \Rightarrow \quad \lim _{x \rightarrow \frac{\pi^{-}}{2}}\left(\frac{8}{7}\right)^{\left(\frac{\tan 8 x}{\tan 7 x}\right)} \quad \text { Let } x=\frac{\pi}{2}-h \\ & \Rightarrow \quad \lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (4 \pi-8 h)}{\tan \left(3 \pi+\frac{\pi}{2}-7 h\right)}}=\lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (-8 h)}{\cot (7 h)}}=\left(\frac{8}{7}\right)^0=1 \\ & \Rightarrow \quad a-8=1 \Rightarrow a=9 \end{aligned}$$</p> <p>$$\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}}(1+|\cot x|)^{\frac{b}{a}|\tan x|}, x=\frac{\pi}{2}+h$$</p> <p>$$\lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h}=\lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h}$$</p> <p>$$\begin{aligned} & =\lim _\limits{h \rightarrow 0}(1-\tan h)^{\left(\frac{-1}{\tan h}\right) \cdot(-\tan h) \cdot\left(\frac{-b}{9} \cot h\right)} \\ & =e^{\frac{b}{9}}=1 \quad \Rightarrow b=0 \\ & \Rightarrow a^2+b^2=81+0=81 \end{aligned}$$</p>