Simplify: sin^2x - cos^2x / 1-tan^2x ?

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Simplify: sin^2x - cos^2x / 1-tan^2x ?

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Answer 1 · 2018-04-12T08:24:14

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

Explanation:

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

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

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

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

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

$= - {cos}^{2} (x)$ $=-cos^2(x)$

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Simplify: sin^2x - cos^2x / 1-tan^2x ?

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