If $\tan \theta=\frac{a}{b}$, then $b \cos 2 \theta+a \sin 2 \theta=$
If $\tan \theta=\frac{a}{b}$, then $b \cos 2 \theta+a \sin 2 \theta=$
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Solution
$\begin{aligned} & b \cos 2 \theta+a \sin 2 \theta=b \times \frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}+a \times \frac{2 \tan \theta}{1+\tan ^2 \theta} \\ & =b \times \frac{1-\frac{a^2}{b^2}}{1+\frac{a^2}{b^2}}+a \times \frac{2 \times \frac{a}{b}}{1+\frac{a^2}{b^2}} \\ & =\frac{b\left(b^2-a^2\right)}{b^2+a^2}+\frac{2 a^2 b}{b^2+a^2} \\ & =\frac{b\left(b^2+a^2\right)}{b^2+a^2}=b\end{aligned}$
Asked in: MHT CET 2022 (06 Aug Shift 1)
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