Question #de16d

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Question #de16d

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Answer 1 · 2016-08-06T08:02:48

We have: $\frac{{cos}^{2} x - {sin}^{2} x}{{cot}^{2} x - {tan}^{2} x}$ $(cos^2x-sin^2x)/(cot^2x-tan^2x)$

Start by writing tan and cot in terms of sin and cos

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

$= \frac{{cos}^{2} x - {sin}^{2} x}{\frac{{cos}^{4} x - {sin}^{4} x}{{sin}^{2} x {cos}^{2} x}} \cdot \frac{{sin}^{2} x {cos}^{2} x}{{sin}^{2} x {cos}^{2} x}$ $=(cos^2x-sin^2x)/((cos^4x-sin^4x)/(sin^2xcos^2x))*(sin^2xcos^2x)/(sin^2xcos^2x)$

$= \frac{{cos}^{4} x {sin}^{2} x - {sin}^{4} x {cos}^{2} x}{{cos}^{4} x - {sin}^{4} x}$ $=(cos^4xsin^2x-sin^4xcos^2x)/(cos^4x-sin^4x)$

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

$=(sin^2xcos^2x)/1$

$=sin^2xcos^2x$

Answer 2 · 2016-08-06T09:29:02

We have:
$LHS=(cos^2x-sin^2x)/(cot^2x-tan^2$
$=(sin^2xcos^2x(cos^2x/(sin^2xcos^2x)-sin^2x/(sin^2xcos^2x)))/(csc^2x-1-sec^2x+1)$

$=((sin^2xcos^2x)(csc^2x-sec^2x))/(csc^2x-sec^2x)$

= $sin^2xcos^2x$

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Question #de16d

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