# How do you prove (1 + tan^2x)/(1-tan^2x) = 1/(cos^2x - sin^2x)?

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Bdub Share
Mar 8, 2018

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#### Explanation:

$L H S : \frac{1 + {\tan}^{2} x}{1 - {\tan}^{2} x}$

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

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

$= \frac{{\cos}^{2} x + {\sin}^{2} x}{\cancel{{\cos}^{2} x}} \cdot \frac{\cancel{{\cos}^{2} x}}{{\cos}^{2} x - {\sin}^{2} x}$

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

$= \frac{1}{{\cos}^{2} x - {\sin}^{2} x}$

$= R H S$

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