What is $\frac{{tan}^{2} θ}{1 - {sin}^{4} θ} - {csc}^{2} θ$ $tan^2theta/(1-sin^4theta)-csc^2theta$ in terms of non-exponential trigonometric functions?

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What is $\frac{{tan}^{2} θ}{1 - {sin}^{4} θ} - {csc}^{2} θ$ $tan^2theta/(1-sin^4theta)-csc^2theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-08-02T13:35:58

$\frac{{tan}^{2} θ}{1 - {sin}^{4} θ} - {csc}^{2} θ$ $tan^2theta/(1-sin^4theta)-csc^2theta$

$= \frac{{sin}^{2} θ}{{cos}^{2} θ} \cdot \frac{1}{(1 - {sin}^{2} θ) (1 + {sin}^{2} θ)} - {csc}^{2} θ$ $=sin^2theta/cos^2theta*1/((1-sin^2theta)(1+sin^2theta))-csc^2theta$

$= \frac{{sin}^{2} θ}{{cos}^{4} θ} \cdot \frac{1}{1 + {sin}^{2} θ} - {csc}^{2} θ$ $=sin^2theta/cos^4theta*1/(1+sin^2theta)-csc^2theta$

$= \frac{8 {sin}^{2} θ}{{(2 {cos}^{2} θ)}^{2}} \cdot \frac{1}{2 + 2 {sin}^{2} θ} - {csc}^{2} θ$ $=(8sin^2theta)/(2cos^2theta)^2*1/(2+2sin^2theta)-csc^2theta$

$= \frac{4 (1 - cos 2 θ)}{{(1 + cos 2 θ)}^{2}} \cdot \frac{1}{2 + 1 - cos 2 θ} - {csc}^{2} θ$ $=(4(1-cos2theta))/(1+cos2theta)^2*1/(2+1-cos2theta)-csc^2theta$

$= \frac{8 (1 - cos 2 θ)}{2 + 4 cos 2 θ + 2 {cos}^{2} 2 θ} \cdot \frac{1}{3 - cos 2 θ} - \frac{2}{2 {sin}^{2} θ}$ $=(8(1-cos2theta))/(2+4cos2theta+2cos^2 2theta)*1/(3-cos2theta)-2/(2sin^2theta)$

$= \frac{8 (1 - cos 2 θ)}{3 + 4 cos 2 θ + cos 4 θ} \cdot \frac{1}{3 - cos 2 θ} - \frac{2}{1 - cos 2 θ}$ $=(8(1-cos2theta))/(3+4cos2theta+cos4theta)*1/(3-cos2theta)-2/(1-cos2theta)$

Answer 2 · 2016-08-02T13:36:30

$\frac{{tan}^{2} θ}{1 - {sin}^{4} θ} - {csc}^{2} θ$ $tan^2theta/(1-sin^4theta)-csc^2theta$

$= \frac{{sin}^{2} θ}{{cos}^{2} θ} \cdot \frac{1}{(1 - {sin}^{2} θ) (1 + {sin}^{2} θ)} - {csc}^{2} θ$ $=sin^2theta/cos^2theta*1/((1-sin^2theta)(1+sin^2theta))-csc^2theta$

$= \frac{{sin}^{2} θ}{{cos}^{4} θ} \cdot \frac{1}{1 + {sin}^{2} θ} - {csc}^{2} θ$ $=sin^2theta/cos^4theta*1/(1+sin^2theta)-csc^2theta$

$= \frac{8 {sin}^{2} θ}{{(2 {cos}^{2} θ)}^{2}} \cdot \frac{1}{2 + 2 {sin}^{2} θ} - {csc}^{2} θ$ $=(8sin^2theta)/(2cos^2theta)^2*1/(2+2sin^2theta)-csc^2theta$

$= \frac{4 (1 - cos 2 θ)}{{(1 + cos 2 θ)}^{2}} \cdot \frac{1}{2 + 1 - cos 2 θ} - {csc}^{2} θ$ $=(4(1-cos2theta))/(1+cos2theta)^2*1/(2+1-cos2theta)-csc^2theta$

$= \frac{8 (1 - cos 2 θ)}{2 + 4 cos 2 θ + 2 {cos}^{2} 2 θ} \cdot \frac{1}{3 - cos 2 θ} - \frac{2}{2 {sin}^{2} θ}$ $=(8(1-cos2theta))/(2+4cos2theta+2cos^2 2theta)*1/(3-cos2theta)-2/(2sin^2theta)$

$= \frac{8 (1 - cos 2 θ)}{3 + 4 cos 2 θ + cos 4 θ} \cdot \frac{1}{3 - cos 2 θ} - \frac{2}{1 - cos 2 θ}$ $=(8(1-cos2theta))/(3+4cos2theta+cos4theta)*1/(3-cos2theta)-2/(1-cos2theta)$

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What is $\frac{{tan}^{2} θ}{1 - {sin}^{4} θ} - {csc}^{2} θ$ $tan^2theta/(1-sin^4theta)-csc^2theta$ in terms of non-exponential trigonometric functions?

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