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$\large\frac{1-\tan^2(45^{\large\circ}-A)}{1+\tan^2(45^{\large\circ}-A)}$ is equal to

$(a)\;\sin 2A\qquad(b)\;\cos 2A\qquad(c)\;\tan 2A\qquad(d)\;\cot 2A$

1 Answer

$\large\frac{1-\tan^2(45^{\large\circ}-A)}{1+\tan^2(45^{\large\circ}-A)}$
$\Rightarrow \cos(90^{\large\circ}-2A)$
$\Rightarrow \sin 2A$
Hence (a) is the correct answer.
answered Oct 9, 2013 by sreemathi.v
 

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