Prove that ${cot}^{2} x - {cos}^{2} x = {cot}^{2} x {cos}^{2} x$ $cot^2x-cos^2x=cot^2xcos^2x$ ?

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Prove that ${cot}^{2} x - {cos}^{2} x = {cot}^{2} x {cos}^{2} x$ $cot^2x-cos^2x=cot^2xcos^2x$ ?

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Answer 1 · 2017-03-22T03:37:43

Please see below.

Explanation:

${cot}^{2} x - {cos}^{2} x$ $cot^2x-cos^2x$

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

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

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

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

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

= ${cot}^{2} x {cos}^{2} x$ $cot^2xcos^2x$

Answer 2 · 2017-03-22T03:38:11

Develop the left side:
$L S = \frac{{cos}^{2} x}{{sin}^{2} x} - {cos}^{2} x = \frac{({cos}^{2} x) (1 - {sin}^{2} x)}{{sin}^{2} x} =$ $LS = (cos^2 x)/(sin^2 x) - cos ^2 x = ((cos^2 x)(1 - sin^2 x))/(sin^2 x) =$
$= \frac{{cos}^{2} x . {cos}^{2} x}{{sin}^{2} x} = {cot}^{2} x . {cos}^{2} x$ $= (cos^2 x.cos^2 x)/(sin^2 x) = cot^2 x.cos^2 x$ Proved.

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Prove that ${cot}^{2} x - {cos}^{2} x = {cot}^{2} x {cos}^{2} x$ $cot^2x-cos^2x=cot^2xcos^2x$ ?

2 Answers

Explanation:

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Prove that ${cot}^{2} x - {cos}^{2} x = {cot}^{2} x {cos}^{2} x$ $cot^2x-cos^2x=cot^2xcos^2x$ ?