The maximum expansion corresponds to \({M_2} \to \infty \).We need to obtain Prandtl-Meyer expansion function  \(v\left( M \right)\).

\(\nu \left( M \right) = \sqrt {\frac{{\gamma + 1}}{{\gamma – 1}}} {\tan ^{ – 1}}\sqrt {\frac{{\gamma – 1}}{{\gamma + 1}}\left( {{M^2} – 1} \right)} – {\tan ^{ – 1}}\sqrt {{M^2} – 1} \)

So

\(\begin{array}{l}\mathop {\lim }\limits_{{M_2} \to \infty } \nu \left( {{M_2}} \right) = \mathop {\lim }\limits_{{M_2} \to \infty } \left\{ {\sqrt {\frac{{\gamma + 1}}{{\gamma – 1}}} {{\tan }^{ – 1}}\sqrt {\frac{{\gamma – 1}}{{\gamma + 1}}\left( {M_2^2 – 1} \right)} – {{\tan }^{ – 1}}\sqrt {M_2^2 – 1} } \right\}\\ = \sqrt {\frac{{\gamma + 1}}{{\gamma – 1}}} \times \frac{\pi }{2} – \frac{\pi }{2} – {\left( {\sqrt {\frac{{\gamma + 1}}{{\gamma – 1}}} – 1} \right)^{\frac{\pi }{2}}}\\ = 2.277\,{\rm{radians}}\\ = {130.45^ \circ }\end{array}\)

We need to  obtain \(v\left( M \right)\) for M₁ =1

\(\begin{array}{l}{\nu _1} = 0,\\{\rm{therefore}}\,{\theta _{\max }} = {\nu _2} – {\nu _1}\\ = 134.45 – 0\\ = {130.45^ \circ }\end{array}\)

Hence maximum deflection angle is \({{{130.45}^ \circ }}\).