find the exact value of csc theta if cot theta = -2 and the terminal side of theta lies in quadrant II (2).
2 answers:
Answer:
Not an answer
Step-by-step explanation:
Did you ever find the rest of the test . I’m really struggling.
let's recall that on the II Quadrant x/cosine is negative whilst y/sine is positive,
also let's recall that the hypotenuse is simply the radius distance and thus is never negative.
![\bf cot(\theta )=\cfrac{\stackrel{adjacent}{-2}}{\stackrel{opposite}{1}}\impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{(-2)^2+1^2}\implies c=\sqrt{5} \\\\[-0.35em] ~\dotfill\\\\ csc(\theta )=\cfrac{\stackrel{hypotenuse}{\sqrt{5}}}{\stackrel{opposite}{1}}\implies csc(\theta )=\sqrt{5}](https://tex.z-dn.net/?f=%5Cbf%20cot%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B-2%7D%7D%7B%5Cstackrel%7Bopposite%7D%7B1%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20hypotenuse%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7B%28-2%29%5E2%2B1%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B5%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20csc%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bhypotenuse%7D%7B%5Csqrt%7B5%7D%7D%7D%7B%5Cstackrel%7Bopposite%7D%7B1%7D%7D%5Cimplies%20csc%28%5Ctheta%20%29%3D%5Csqrt%7B5%7D)
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