$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}{\left(\frac{3}{2}\right)}^{(\cot 3\mathrm{x})/(\cot 2\mathrm{x})},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<\mathrm{x}<\frac{\mathrm{\pi }}{2}\\ \mathrm{b}+3,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}=\frac{\mathrm{\pi }}{2}\\ (1+|\cot \mathrm{x}|{)}^{\left(\mathrm{a}\right|\tan \mathrm{x}\left|\right)/\mathrm{b}},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{\mathrm{\pi }}{2}<\mathrm{x}<\mathrm{\pi }\end{array}\right.$ is continuous at $\mathrm{x}=\mathrm{\pi }/2$. Then
 

$\begin{array}{l}\mathrm{f}\left(\frac{{\mathrm{\pi }}^{-}}{2}\right)=\underset{\mathrm{h}\to 0}{lim} {\left(\frac{3}{2}\right)}^{\cot \left(3\left(\frac{\mathrm{\pi }}{2}-\mathrm{h}\right)\right)/\cot \left(2\left(\frac{\mathrm{\pi }}{2}-\mathrm{h}\right)\right)}\\ =\underset{\mathrm{h}\to 0}{lim} {\left(\frac{3}{2}\right)}^{\frac{\tan 3\mathrm{h}}{\cot 2\mathrm{h}}}\\ =\underset{\mathrm{h}\to 0}{lim} {\left(\frac{3}{2}\right)}^{-(\tan 3\mathrm{h})(\tan 2\mathrm{h})}\\ =1\end{array}$
$\begin{array}{l}\mathrm{f}\left(\frac{{\mathrm{\pi }}^{+}}{2}\right)\left.=\underset{\mathrm{h}\to 0}{lim} \left[1+\left|\cot \left(\frac{\mathrm{\pi }}{2}+\mathrm{h}\right)\right|\right]\left[\mathrm{a}\mid \tan \left(\frac{\mathrm{\pi }}{2}+\mathrm{h}\right)\right]\right]\mathrm{b}\\ =\underset{\mathrm{h}\to 0}{lim} (1+\tan \mathrm{h}{)}^{\frac{\mathrm{acot}\mathrm{h}}{\mathrm{b}}}\\ ={\mathrm{e}}^{\underset{\mathrm{h}\to 0}{lim} (1+\tan \mathrm{h}-1)\frac{\mathrm{acot}\mathrm{h}}{\mathrm{b}}}\\ ={\mathrm{e}}^{{\mathrm{a}}^{\mathrm{\prime }}\mathrm{b}}\end{array}$
Also, $\mathrm{f}\left(\frac{\mathrm{\pi }}{2}\right)=\mathrm{b}+3$
f (x) is continuous at $\mathrm{x}=\mathrm{\pi }/2$. Therefore,
$\mathrm{1=}\mathrm{b}+3={\mathrm{e}}^{\mathrm{a}/\mathrm{b}} \mathrm{or} \mathrm{b}=-2 \mathrm{and} \mathrm{a}=0$