What is $cot θ + tan θ \cdot sec θ$ $cot theta + tantheta*sectheta$ in terms of $sin θ$ $sintheta$ ?

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What is $cot θ + tan θ \cdot sec θ$ $cot theta + tantheta*sectheta$ in terms of $sin θ$ $sintheta$ ?

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Answer 1 · 2016-01-27T11:35:26

$\frac{{(1 - {sin}^{2} θ)}^{\frac{3}{2}} + {sin}^{2} θ}{sin θ (1 - {sin}^{2} θ)}$ $((1-sin^2theta)^(3/2) +sin^2theta)/(sintheta(1-sin^2theta)$

Explanation:

You need to use the facts that $cot θ = \frac{1}{tan θ}$ $cot theta = 1/tan theta$ and that $tan θ = \frac{sin θ}{cos θ}$ $tan theta = sin theta/cos theta$

Then
$cot θ + tan θ \cdot sec θ = \frac{cos θ}{sin θ} + \frac{sin θ}{cos θ} \cdot \frac{1}{cos θ}$ $cot theta +tan theta * sec theta = cos theta/sin theta + sin theta / cos theta * 1/cos theta$

We know that ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta + cos^2theta = 1$ so

${cos}^{2} θ = 1 - {sin}^{2} θ$ $cos^2theta = 1 - sin^2theta$ and

$cos θ = \sqrt{1 - {sin}^{2} θ}$ $costheta = sqrt(1-sin^2theta)$

Substituting these into the expression gives

$\frac{\sqrt{1 - {sin}^{2} θ}}{sin θ} + \frac{sin θ}{1 - {sin}^{2} θ}$ $sqrt(1-sin^2theta)/sintheta +sintheta/(1-sin^2theta)$

$= \frac{{(1 - {sin}^{2} θ)}^{\frac{3}{2}} + {sin}^{2} θ}{sin θ (1 - {sin}^{2} θ)}$ $=((1-sin^2theta)^(3/2) +sin^2theta)/(sintheta(1-sin^2theta)$

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What is $cot θ + tan θ \cdot sec θ$ $cot theta + tantheta*sectheta$ in terms of $sin θ$ $sintheta$ ?

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