If \(\mathrm{a} \sin \theta=\mathrm{b} \cos \theta\), where \(\mathrm{a}, \mathrm{b} \neq 0\), then \(\mathrm{a} \cos 2 \theta+\mathrm{b} \sin 2 \theta=\)

\( a \sin \theta=b \cos \theta \Rightarrow \tan \theta=\frac{b}{a} \)
\( a \cos 2 \theta+b \sin 2 \theta \)
\( =a\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)+b\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) \)
\( =a\left[\frac{1-\left(\frac{b^2}{a^2}\right)}{1+\left(\frac{b^2}{a^2}\right)}\right]+b\left[\frac{2\left(\frac{b}{a}\right)}{1+\left(\frac{b^2}{a^2}\right)}\right]=a\left(\frac{a^2-b^2}{a^2+b^2}\right)+\) \(b\left(\frac{2 b}{a} \times \frac{a^2}{a^2+b^2}\right) \)
\( =\frac{a\left(a^2-b^2\right)}{a^2+b^2}+\frac{b(2 a b)}{a^2+b^2}=\frac{a^3-a b^2+2 a b^2}{a^2+b^2} \)
\( =\frac{a^3+a b^2}{a^2+b^2}=\frac{a\left(a^2+b^2\right)}{a^2+b^2}=a\)