Question #fbd74

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Question #fbd74

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Answer 1 · 2017-03-19T03:09:52

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

Work on left side:

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

${cos}^{2} x$ $cos^2x$ cancels, leaving:

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

Change tan to $\frac{sin}{cos}$ $sin/cos$ :

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

Multiply $(\frac{{sin}^{2} x}{1})$ $(Sin^2x/1)$ by $(\frac{{cos}^{2} x}{{cos}^{2} x})$ $(cos^2x/cos^2x)$ to get a common denominator:

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

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

${cos}^{2} x$ $cos^2x$ cancels:

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

Factor the numerator:

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

${sin}^{2} x$ $sin^2x$ cancels:

$1 - {cos}^{2} x$ $1-cos^2x$ which is also ${sin}^{2} x$ $sin^2x$ = right side

Answer 2 · 2017-03-19T04:45:35

$L H S = \frac{{cos}^{2} x + {tan}^{2} x - 1}{{tan}^{2} x}$ $LHS=(cos^2x + tan^2x - 1)/tan^2x$

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

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

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

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

$= 1 - {cos}^{2} x = {sin}^{2} x = R H S$ $=1 - cos^2x=sin^2x=RHS$

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