If tan θ=

Given $\tan \theta=\frac{a}{b}$

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

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

$\left[\right.$ using identities : $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$ and $\left.\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right]$

$=\frac{2 a \tan \theta+b-b \tan ^{2} \theta}{1+\tan ^{2} \theta}$

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

$=\frac{\frac{2 a^{2}}{b}+b-\frac{a^{2}}{b}}{\frac{b^{2}+a^{2}}{b^{2}}}$

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

$=\frac{a^{2}+b^{2}}{b} \times \frac{b^{2}}{a^{2}+b^{2}}$

$=b$

Therefore,

$a \sin 2 \theta+b \cos 2 \theta=b$