How do you express $tan θ - cot θ$ $tan theta - cot theta$ in terms of $cos θ$ $cos theta$ ?

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How do you express $tan θ - cot θ$ $tan theta - cot theta$ in terms of $cos θ$ $cos theta$ ?

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Answer 1 · 2016-03-25T02:32:06

We must use the quotient identities, $tan θ = \frac{sin θ}{cos θ}$ $tantheta = sintheta/costheta$ and $cot θ = \frac{cos θ}{sin θ}$ $cottheta = costheta/sintheta$

Explanation:

= $\frac{sin θ}{cos θ} - \frac{cos θ}{sin θ}$ $sintheta/costheta - costheta/sintheta$

= $\frac{{sin}^{2} θ}{cos θ sin θ} - \frac{{cos}^{2} θ}{cos θ sin θ}$ $sin^2theta/(costhetasintheta) - cos^2theta/(costhetasintheta)$

Use the pythagorean identity ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta + cos^2theta = 1$ to simplify further.

$= \frac{1 - {cos}^{2} θ - {cos}^{2} θ}{cos θ sin θ}$ $= (1 - cos^2theta - cos^2theta)/(costhetasintheta)$

$= \frac{1 - 2 {cos}^{2} θ}{cos θ sin θ}$ $= (1 - 2cos^2theta)/(costhetasintheta)$

This is simplest form; we can't get rid of the sin (pardon the unintended pun!).

Hopefully this helps!

Answer 2 · 2016-03-28T13:24:35

$\frac{\sqrt{1 - {cos}^{2} θ}}{cos θ} - \frac{cos θ}{\sqrt{1 - {cos}^{2} θ}}$ $sqrt(1-cos^2theta)/costheta-costheta/sqrt(1-cos^2theta)$

Explanation:

Express first in terms of $sin θ$ $sintheta$ and $cos θ$ $costheta$ .

$= \frac{sin θ}{cos θ} - \frac{cos θ}{sin θ}$ $=sintheta/costheta-costheta/sintheta$

Now, to turn the $sin θ$ $sintheta$ terms into $cos θ$ $costheta$ terms, use the Pythagorean identity:

${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$

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

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

Substitute this into the expression we originally made:

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

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How do you express $tan θ - cot θ$ $tan theta - cot theta$ in terms of $cos θ$ $cos theta$ ?

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