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If $\large\frac{1+\sin 2x}{1-\sin 2x}$$=\cot^2(a+x)$ $\forall x\in R\sim \big(n\pi+\large\frac{\pi}{4}\big)$$ n\in N$ then $'a'$ is equal to


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$\large\frac{1+\sin 2x}{1-\sin 2x}=\frac{(\sin x+\cos x)^2}{(\sin x-\cos x)^2}$
$\qquad\qquad=\big(\large\frac{1+\cot x}{1-\cot x}\big)^2$
$\Rightarrow \big(\large\frac{1+\tan x}{1-\tan x}\big)^2$$=(\tan(\large\frac{\pi}{4}$$+x))^2$
Hence (c) is the correct answer.
answered Oct 7, 2013 by sreemathi.v

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