# If $\tan \theta=\large\frac{a}{b}$,then $b\cos 2\theta+a\sin 2\theta$ is equal to

$\begin{array}{1 1}(A)\;a&(B)\;b\\(C)\;\large\frac{a}{b}&(D)\;\text{None}\end{array}$

Toolbox:
• $\cos 2\theta=\large\frac{1-\tan^22\theta}{1+\tan^2\theta}$
• $\sin 2\theta=\large\frac{2\tan \theta}{1+\tan^2\theta}$
Given :
$b\cos 2\theta+a\sin 2\theta$
$b\times \big(\large\frac{1-\tan^2\theta}{1+\tan^2\theta}\big)$$+a\big(\large\frac{2\tan \theta}{1+\tan^2\theta}\big) b\times \big(\large\frac{1-\Large\frac{a^2}{b^2}}{1+\Large\frac{a^2}{b^2}}\big)+$$a\big(\large\frac{2\times (\Large\frac{a}{b})}{1+\Large\frac{a^2}{b^2}}\big)$
$\large\frac{b(\Large\frac{b^2-a^2}{b^2})}{\Large\frac{b^2+a^2}{b^2}}+\frac{\Large\frac{2a^2}{b}}{\Large\frac{b^2+a^2}{b^2}}$
$\Rightarrow \large\frac{b(b^2-a^2)}{b^2+a^2}+\frac{2a^2b}{b^2+a^2}$
$\Rightarrow \large\frac{b(b^2-a^2)+2a^2b}{b^2+a^2}$
$\Rightarrow \large\frac{b(b^2-a^2+2a^2)}{b^2+a^2}$
$\Rightarrow \large\frac{b(b^2+a^2)}{b^2+a^2}$
$\Rightarrow b$
Hence (B) is the correct answer.