Question #4352a

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Question #4352a

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Answer 1 · 2017-03-26T18:34:26

$1$ $1$

Explanation:

If $3 {sin}^{2} x - {sin}^{4} x = 1$ $3sin^2x-sin^4x=1$ then ${sin}^{2} x = \frac{3 - \sqrt{5}}{2}$ $sin^2x=(3-sqrt(5))/2$

and

${cos}^{2} x = 1 - {sin}^{2} x = 1 - \frac{3 - \sqrt{5}}{2}$ $cos^2x=1-sin^2x=1-(3-sqrt(5))/2$

then

${tan}^{2} x + {tan}^{4} x = \frac{\frac{3 - \sqrt{5}}{2}}{1 - \frac{3 - \sqrt{5}}{2}} + {⎛ ⎜ ⎝ \frac{\frac{3 - \sqrt{5}}{2}}{1 - \frac{3 - \sqrt{5}}{2}} ⎞ ⎟ ⎠}^{2} = 1$ $tan^2x+tan^4x =((3-sqrt(5))/2)/(1-(3-sqrt(5))/2)+(((3-sqrt(5))/2)/(1-(3-sqrt(5))/2))^2=1$

Answer 2 · 2017-03-27T03:52:21

$3 - {sin}^{2} x$ $3 - sin^2 x$

Explanation:

Develop the left side of the equation:
$3 {sin}^{2} x - {sin}^{4} x = 1$ $3sin^2 x - sin^4 x = 1$
${sin}^{2} x (3 - {sin}^{2} x) = 1$ $sin^2 x(3 - sin^2 x) = 1$
From this equation we get:
$\frac{1}{{sin}^{2} x} = 3 - {sin}^{2} x$ $1/sin^2 x = 3 - sin^2 x$ (1)

Develop the right side:
${tan}^{2} x + {tan}^{4} x = {tan}^{2} x (1 + {tan}^{2} x) = {tan}^{2} x (\frac{1}{{cos}^{2} x}) =$ $tan^2 x + tan^4 x = tan^2 x(1 + tan^2 x) = tan^2 x(1/cos^2 x) =$
$= \frac{{cos}^{2} x}{{sin}^{2} x} (\frac{1}{{cos}^{2} x}) = \frac{1}{{sin}^{2} x} (2)$ $= (cos^2 x)/(sin^2 x)(1/(cos^2 x)) = 1/(sin^2 x) (2)$
Compare (1) and (2) -->
${tan}^{2} x + {tan}^{4} x = 3 - {sin}^{2} x$ $tan^2 x + tan^4 x = 3 - sin^2 x$

Answer 3 · 2017-03-27T15:10:25

$3 {sin}^{2} x - {sin}^{4} x = 1$ $3sin^2x-sin^4x=1$

Dividing bothsides by ${cos}^{4} x$ $cos^4x$

$\Rightarrow \frac{3 {sin}^{2} x}{{cos}^{4} x} - \frac{{sin}^{4} x}{{cos}^{4} x} = \frac{1}{{cos}^{4} x}$ $=>(3sin^2x)/cos^4x-sin^4x/cos^4x=1/cos^4x$

$\Rightarrow 3 {tan}^{2} x {sec}^{2} x - {tan}^{4} x = {sec}^{4} x$ $=>3tan^2xsec^2x-tan^4x=sec^4x$

$\Rightarrow 3 {tan}^{2} x (1 + {tan}^{2} x) - {tan}^{4} x = {(1 + {tan}^{2} x)}^{2}$ $=>3tan^2x(1+tan^2x)-tan^4x=(1+tan^2x)^2$

$\Rightarrow 3 {tan}^{2} x + 3 {tan}^{4} x - {tan}^{4} x = 1 + 2 {tan}^{2} x + {tan}^{4} x$ $=>3tan^2x+3tan^4x-tan^4x=1+2tan^2x+tan^4x$

$\Rightarrow 3 {tan}^{2} x - 2 {tan}^{2} x + 3 {tan}^{4} x - {tan}^{4} x - {tan}^{4} x = 1$ $=>3tan^2x-2tan^2x+3tan^4x-tan^4x-tan^4x=1$

$\Rightarrow {tan}^{2} x + {tan}^{4} x = 1$ $=>tan^2x+tan^4x=1$

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Question #4352a

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