Solution: \(\displaystyle{\cos{{\left({\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}\right)}}}\)

Let \(\displaystyle{\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}={x}\Rightarrow{\cos{{\left({x}\right)}}}={\cos{{\left({\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}\right)}}}\)

\(\displaystyle\Rightarrow{\frac{{{v}}}{{{a}}}}={\tan{{x}}}\)

\(\displaystyle\Rightarrow{\frac{{{v}}}{{{a}}}}={\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}\)

\(\displaystyle\Rightarrow\text{Squaring both side}\)

\(\displaystyle{\frac{{{v}^{{2}}}}{{{a}^{{2}}}}}={\frac{{{{\sin}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{v}^{{2}}}}{{{a}^{{2}}}}}+{1}={\frac{{{{\sin}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}+{1}\)

\(\displaystyle\Rightarrow{\frac{{{v}^{{2}}+{a}^{{2}}}}{{{a}^{{2}}}}}={\frac{{{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}\)

\(\displaystyle\therefore{\frac{{{v}^{{2}}+{a}^{{2}}}}{{{a}^{{2}}}}}={\frac{{{1}}}{{{{\cos}^{{2}}{x}}}}}\)

\(\displaystyle\Rightarrow{\cos{{x}}}=\sqrt{{{\frac{{{a}^{{2}}}}{{{v}^{{2}}+{a}^{{2}}}}}}}\)

\(\displaystyle\therefore{\cos{{\left({\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}\right)}}}=\sqrt{{{\frac{{{1}}}{{{\left({\frac{{{v}}}{{{a}}}}\right)}^{{2}}+{1}}}}}}\)

Let \(\displaystyle{\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}={x}\Rightarrow{\cos{{\left({x}\right)}}}={\cos{{\left({\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}\right)}}}\)

\(\displaystyle\Rightarrow{\frac{{{v}}}{{{a}}}}={\tan{{x}}}\)

\(\displaystyle\Rightarrow{\frac{{{v}}}{{{a}}}}={\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}\)

\(\displaystyle\Rightarrow\text{Squaring both side}\)

\(\displaystyle{\frac{{{v}^{{2}}}}{{{a}^{{2}}}}}={\frac{{{{\sin}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{v}^{{2}}}}{{{a}^{{2}}}}}+{1}={\frac{{{{\sin}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}+{1}\)

\(\displaystyle\Rightarrow{\frac{{{v}^{{2}}+{a}^{{2}}}}{{{a}^{{2}}}}}={\frac{{{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}\)

\(\displaystyle\therefore{\frac{{{v}^{{2}}+{a}^{{2}}}}{{{a}^{{2}}}}}={\frac{{{1}}}{{{{\cos}^{{2}}{x}}}}}\)

\(\displaystyle\Rightarrow{\cos{{x}}}=\sqrt{{{\frac{{{a}^{{2}}}}{{{v}^{{2}}+{a}^{{2}}}}}}}\)

\(\displaystyle\therefore{\cos{{\left({\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}\right)}}}=\sqrt{{{\frac{{{1}}}{{{\left({\frac{{{v}}}{{{a}}}}\right)}^{{2}}+{1}}}}}}\)