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

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

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Answer 1 · 2018-04-21T03:13:00

$L H S = \frac{1 + {tan}^{2} x}{1 - {tan}^{2} x}$ $LHS =(1 + tan^2x)/(1-tan^2x)$

$= \frac{1 + \frac{{sin}^{2} x}{{cos}^{2} x}}{1 - \frac{{sin}^{2} x}{{cos}^{2} x}}$ $=(1+sin^2x/cos^2x)/(1-sin^2x/cos^2x)$

$= \frac{\frac{{cos}^{2} x + {sin}^{2} x}{{cos}^{2} x}}{\frac{{cos}^{2} x - {sin}^{2} x}{{cos}^{2} x}}$ $=((cos^2x+sin^2x)/cos^2x)/((cos^2x-sin^2x)/cos^2x)$

$= \frac{{sin}^{2} x + {cos}^{2} x}{{cos}^{2} x - {sin}^{2} x} = R H S$ $= (sin^2x+cos^2x)/(cos^2x - sin^2x) =RHS$

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

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