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

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

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Answer 1 · 2015-05-31T02:37:43

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

= $\frac{{sin}^{2} x}{{cos}^{2} x} = {tan}^{2} x$ $sin^2 x/cos^2 x = tan^2 x$

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

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How do you prove tan2(x)sec(x)−1=(sec(x)+1)tan^2(x) / (sec(x) - 1) = (sec(x) + 1)?

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