Question #0280a

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Question #0280a

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Answer 1 · 2016-10-01T16:30:51

$sin (θ) = \frac{6 \sqrt{52}}{52}$ $sin(theta) = (6sqrt52)/52$ , $cos (θ) = \frac{- 4 \sqrt{52}}{52}$ $cos(theta) = (-4sqrt52)/52$ , $sec (θ) = - \frac{\sqrt{52}}{4}$ $sec(theta) = -sqrt52/4$ , $csc = \frac{\sqrt{52}}{6}$ $csc = sqrt(52)/6$ and $cot (θ) = - \frac{4}{6}$ $cot(theta) = -4/6$

Explanation:

The cotangent is the reciprocal of the tangent:

$cot (θ) = - \frac{4}{6}$ $cot(theta) = -4/6$

Use $1 + {tan}^{2} (θ) = {sec}^{2} (θ)$ $1 + tan^2(theta) = sec^2(theta)$ to obtain the secant and the cosine; we know that both the secant and the cosine must be negative because we are given $sin (θ) > 0$ $sin(theta) > 0$

$1 + {(\frac{- 6}{4})}^{2} = {sec}^{2} (θ)$ $1 + ((-6)/4)^2 = sec^2(theta)$

${sec}^{2} (θ) = \frac{16}{16} + \frac{36}{16}$ $sec^2(theta) = 16/16 + 36/16$

${sec}^{2} (θ) = \frac{52}{16}$ $sec^2(theta) = 52/16$

$sec (θ) = - \frac{\sqrt{52}}{4}$ $sec(theta) = -sqrt52/4$

$cos (θ) = \frac{- 4 \sqrt{52}}{52}$ $cos(theta) = (-4sqrt52)/52$

$sin (θ) = (- \frac{6}{4}) cos (θ)$ $sin(theta) = (-6/4)cos(theta)$

$sin (θ) = \frac{6 \sqrt{52}}{52}$ $sin(theta) = (6sqrt52)/52$

$csc = \frac{\sqrt{52}}{6}$ $csc = sqrt(52)/6$

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Question #0280a

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