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

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

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Answer 1 · 2017-03-22T03:37:43

Please see below.

Explanation:

$cot^2x-cos^2x$

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

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

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

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

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

= $cot^2xcos^2x$

Answer 2 · 2017-03-22T03:38:11

Develop the left side:
$LS = (cos^2 x)/(sin^2 x) - cos ^2 x = ((cos^2 x)(1 - sin^2 x))/(sin^2 x) =$
$= (cos^2 x.cos^2 x)/(sin^2 x) = cot^2 x.cos^2 x$ Proved.

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

2 Answers

Explanation:

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