How do you prove ${sin}^{2} x + \frac{{tan}^{2} x}{1 - {tan}^{2} x} = \frac{1}{{cos}^{2} x - {sin}^{2} x} - {cos}^{2} x$ $sin^2x + tan^2x/(1-tan^2x) = 1/(cos^2x - sin^2x) -cos^2x$ ?

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How do you prove ${sin}^{2} x + \frac{{tan}^{2} x}{1 - {tan}^{2} x} = \frac{1}{{cos}^{2} x - {sin}^{2} x} - {cos}^{2} x$ $sin^2x + tan^2x/(1-tan^2x) = 1/(cos^2x - sin^2x) -cos^2x$ ?

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Answer 1 · 2018-03-28T13:18:57

Please refer to the Explanation.

Explanation:

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

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

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

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

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

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

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

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

Please check the Problem.

Answer 2 · 2018-03-28T13:26:24

Please see below.
${sin}^{2} x + \frac{2 {tan}^{2} x}{1 - {tan}^{2} x} = \frac{1}{{cos}^{2} x - {sin}^{2} x} - {cos}^{2} x$ $sin^2x +(color(red)(2)tan^2x)/(1-tan^2x) = 1/(cos^2x - sin^2x) -cos^2x$ ?

Explanation:

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

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

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

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

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

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

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

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

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

$= R H S$ $=RHS$

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How do you prove ${sin}^{2} x + \frac{{tan}^{2} x}{1 - {tan}^{2} x} = \frac{1}{{cos}^{2} x - {sin}^{2} x} - {cos}^{2} x$ $sin^2x + tan^2x/(1-tan^2x) = 1/(cos^2x - sin^2x) -cos^2x$ ?

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Explanation:

Explanation:

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Explanation:

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How do you prove ${sin}^{2} x + \frac{{tan}^{2} x}{1 - {tan}^{2} x} = \frac{1}{{cos}^{2} x - {sin}^{2} x} - {cos}^{2} x$ $sin^2x + tan^2x/(1-tan^2x) = 1/(cos^2x - sin^2x) -cos^2x$ ?