If $\tan x=\frac{a}{b}$, then $b \cos 2 x+a \sin 2 x$ is equal to
(a) $a$
(b) $b$
(c) $\frac{a}{b}$
(d) $\frac{b}{a}$
Given: $\tan x=\frac{a}{b}$
Now,
$b \cos 2 x+a \sin 2 x$
$=b\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)+a\left(\frac{2 \tan x}{1+\tan ^{2} x}\right)$
$=b\left(\frac{1-\frac{a^{2}}{b^{2}}}{1+\frac{a^{2}}{b^{2}}}\right)+a\left(\frac{2 \times \frac{a}{b}}{1+\frac{a^{2}}{b^{2}}}\right)$
$=\frac{b\left(b^{2}-a^{2}\right)}{a^{2}+b^{2}}+\frac{2 a^{2} b}{a^{2}+b^{2}}$
$=\frac{b^{3}-a^{2} b+2 a^{2} b}{a^{2}+b^{2}}$
$=\frac{b^{3}+a^{2} b}{a^{2}+b^{2}}$
$=\frac{b\left(b^{2}+a^{2}\right)}{a^{2}+b^{2}}$
$=b$
Hence, the correct answer is option B.
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