Question #04652

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Question #04652

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Answer 1 · 2017-02-07T22:10:23

see below

Explanation:

Left Hand Side:

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

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

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

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

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

$=cancel(sin^2x-cos^2x)/cos^2x *sin^2x/(-cancel(sin^2x-cos^2x)$

$=sin^2x/(-cos^2x)$

$=(1-cos^2x)/(-cos^2x)$

$=1/(-cos^2x) -cos^2x/(-cos^2x)$

$=1/(-cos^2x)+cancel(cos^2x)/cancel(cos^2x)$

$=-1/cos^2x +1$

$=1-1/cos^2x$

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Question #04652

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