How do you prove $\frac{cos (2 x)}{1 - sin 2 x} = \frac{1 + tan 2 x}{1 - tan 2 x}$ $cos (2x) / (1-sin 2x) = (1 + tan 2x) / (1-tan 2x)$ ?

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How do you prove $\frac{cos (2 x)}{1 - sin 2 x} = \frac{1 + tan 2 x}{1 - tan 2 x}$ $cos (2x) / (1-sin 2x) = (1 + tan 2x) / (1-tan 2x)$ ?

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Answer 1 · 2015-08-12T15:37:11

$\frac{cos 2 x}{1 - sin 2 x} \neq \frac{1 + tan 2 x}{1 - tan 2 x}$ $(cos 2x)/(1-sin 2x) ne (1+tan 2x)/(1-tan 2x)$

Explanation:

Assume that $\frac{cos 2 x}{1 - sin 2 x} = \frac{1 + tan 2 x}{1 - tan 2 x}$ $(cos 2x)/(1-sin 2x) = (1+tan 2x)/(1-tan 2x)$ is true for all $x$ $x$

Taking $2 x = \frac{π}{4}$ $2x = pi/4$

LHS: $\frac{cos (\frac{π}{4})}{1 - sin (\frac{π}{4})} = \frac{\sqrt{2}}{2 - \sqrt{2}}$ $cos(pi/4)/(1-sin(pi/4)) = sqrt(2)/(2-sqrt(2))$
RHS: $lim_(2x rarr pi/4) (1+tan(2x))/(1-tan(2x)) = oo$

Therefore $(cos 2x)/(1-sin 2x) ne (1+tan 2x)/(1-tan 2x)$

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How do you prove $\frac{cos (2 x)}{1 - sin 2 x} = \frac{1 + tan 2 x}{1 - tan 2 x}$ $cos (2x) / (1-sin 2x) = (1 + tan 2x) / (1-tan 2x)$ ?

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Explanation:

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How do you prove cos (2x) / (1-sin 2x) = (1 + tan 2x) / (1-tan 2x)cos(2x)1−sin2x=1+tan2x1−tan2xcos (2x) / (1-sin 2x) = (1 + tan 2x) / (1-tan 2x)?

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Explanation:

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How do you prove $\frac{cos (2 x)}{1 - sin 2 x} = \frac{1 + tan 2 x}{1 - tan 2 x}$ $cos (2x) / (1-sin 2x) = (1 + tan 2x) / (1-tan 2x)$ ?