# Express [(1-sin2x)^0.5]÷[(1+sin2x)^0.5] in terms of tanx using the half angle t formula?

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#### Explanation:

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Feb 15, 2018

$\implies \frac{1 - \tan x}{1 + \tan x}$

#### Explanation:

$\sin \left(2 x\right) = \frac{2 \tan x}{1 + {\tan}^{2} x}$

$\sqrt{\frac{1 - \sin \left(2 x\right)}{1 + \sin \left(2 x\right)}}$

=> sqrt((1- ((2tan x) / (1+tan^2 x)))/ (1+ ((2 tan x) / (1+tan^2theta))

$\implies \sqrt{\frac{\frac{1 + {\tan}^{2} x - 2 \tan x}{\cancel{1 + {\tan}^{2} x}}}{\frac{1 + {\tan}^{2} x - 2 \tan x}{\cancel{1 + {\tan}^{2} x}}}}$

$\implies \sqrt{{\left(1 - \tan x\right)}^{2} / {\left(1 + \tan x\right)}^{2}}$

$\implies {\left({\left(\frac{1 - \tan x}{1 + \tan x}\right)}^{\cancel{2}}\right)}^{\cancel{\frac{1}{2}}}$

$\implies \frac{1 - \tan x}{1 + \tan x}$

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