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If $\tan x=\large\frac{b}{a}$ then the value of $a\cos 2x+b\sin 2x$ is

$(a)\;a\qquad(b)\;a-b\qquad(c)\;a+b\qquad(d)\;b$

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1 Answer

$a\cos 2x+b\sin 2x=a\large\frac{1-\tan^2 x}{1+\tan^2 x}+\frac{b.2\tan x}{1+\tan^2 x}$
$\qquad\quad\qquad\qquad=a\large\frac{1-b^2/a^2}{1+b^2/a^2}+\frac{b.2b/a}{1+b^2/a^2}$
$\qquad\quad\qquad\qquad=a\bigg[\large\frac{a^2-b^2}{a^2+b^2}\bigg]+\frac{2ab^2}{a^2+b^2}$
$\qquad\quad\qquad\qquad=\bigg(\large\frac{a}{a^2+b^2}\bigg)$$(a^2-b^2+2b^2)$
$\qquad\quad\qquad\qquad=a$
Hence (a) is the correct option.
answered Oct 9, 2013 by sreemathi.v
 

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