(a) Given that $\tan \theta = \frac{b}{a}$.

Now, $a\cos 2\theta + b\sin 2\theta $

$= a\left( {\frac{{1 – {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) + b\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)$

Putting $\tan \theta = \frac{b}{a}$, we get

$ = a\left( {\frac{{1 – \frac{{{b^2}}}{{{a^2}}}}}{{1 + \frac{{{b^2}}}{{{a^2}}}}}} \right) + b\left( {\frac{{2\frac{b}{a}}}{{1 + \frac{{{b^2}}}{{{a^2}}}}}} \right) $

$= a\left( {\frac{{{a^2} – {b^2}}}{{{a^2} + {b^2}}}} \right) + b\left( {\frac{{2ba}}{{{a^2} + {b^2}}}} \right)$

$ = \frac{1}{{({a^2} + {b^2})}}\{ {a^3} – a{b^2} + 2a{b^2}\} $

$= \frac{{a({a^2} + {b^2})}}{{{a^2} + {b^2}}} = a$.