How do you express $sin θ - cot θ + {tan}^{2} θ$ $sin theta - cot theta + tan^2 theta$ in terms of $cos θ$ $cos theta$ ?

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How do you express $sin θ - cot θ + {tan}^{2} θ$ $sin theta - cot theta + tan^2 theta$ in terms of $cos θ$ $cos theta$ ?

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Answer 1 · 2016-06-07T16:58:56

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

Explanation:

Only Pythagotean identity is used here:
${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2 theta+cos^2 theta=1$

$sin θ - cot θ + {tan}^{2} θ = sin θ - \frac{cos θ}{sin θ} + \frac{{sin}^{2} θ}{{cos}^{2} θ} = \frac{{sin}^{2} θ - cos θ}{sin θ} + \frac{1 - {cos}^{2} θ}{{cos}^{2} θ} = \frac{1 - cos θ - {cos}^{2} θ}{\sqrt{1 - {cos}^{2} θ}} + \frac{1}{{cos}^{2} θ} - 1$ $sin theta-cot theta +tan^2 theta=sin theta-cos theta/sin theta+sin^2 theta/cos^2 theta=(sin^2 theta-cos theta)/sin theta+(1-cos^2 theta)/cos^2 theta=(1-cos theta-cos^2 theta)/sqrt(1-cos^2 theta)+1/cos^2 theta-1$

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How do you express $sin θ - cot θ + {tan}^{2} θ$ $sin theta - cot theta + tan^2 theta$ in terms of $cos θ$ $cos theta$ ?

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How do you express $sin θ - cot θ + {tan}^{2} θ$ $sin theta - cot theta + tan^2 theta$ in terms of $cos θ$ $cos theta$ ?