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

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Answer 1 · 2016-10-03T18:45:46

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$\frac{1 - {tan}^{x}}{1 - {cot}^{2} x} = 1 - {sec}^{2} x$ $(1-tan^x)/(1-cot^2x) = 1-sec^2x$

Left Side $= \frac{1 - {tan}^{2} x}{1 - {cot}^{2} x}$ $=(1-tan^2x)/(1-cot^2x)$

$= \frac{1 - {tan}^{2} x}{1 - (\frac{1}{{tan}^{2} x})}$ $=(1-tan^2x)/(1-(1/tan^2x))$

$= \frac{1 - {tan}^{2} x}{\frac{{tan}^{2} x - 1}{{tan}^{2} x}}$ $=(1-tan^2x)/((tan^2x-1)/tan^2x)$

$= (1 - {tan}^{2} x) \cdot (\frac{{tan}^{2} x}{{tan}^{2} x - 1})$ $=(1-tan^2x) * (tan^2x/(tan^2x-1))$

$= (1 - {tan}^{2} x) \cdot (\frac{{tan}^{2} x}{- (1 - {tan}^{2} x)})$ $=(1-tan^2x) * (tan^2x/-(1-tan^2x))$

$= - {tan}^{2} x$ $=-tan^2x$

$= - ({sec}^{2} x - 1)$ $=-(sec^2x-1)$

$= - {sec}^{2} x + 1$ $=-sec^2x+1$

$= 1 - {sec}^{2} x$ $=1-sec^2x$

$=$ $=$ Right Side

How do you verify (1-tan^2x)/(1-cot^2x) = 1-sec^2x1−tan2x1−cot2x=1−sec2x(1-tan^2x)/(1-cot^2x) = 1-sec^2x?