Question

find the exact value of csc theta if cot theta = -2 and the terminal side of theta lies in quadrant II (2).

Answer 1

Answer:

Not an answer

Step-by-step explanation:

Did you ever find the rest of the test . I’m really struggling.

Answer 2

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.

$\bf cot(\theta )=\cfrac{\stackrel{adjacent}{-2}}{\stackrel{opposite}{1}}\impliedby \textit{let's find the hypotenuse} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{(-2)^2+1^2}\implies c=\sqrt{5} \\\\[-0.35em] ~\dotfill\\\\ csc(\theta )=\cfrac{\stackrel{hypotenuse}{\sqrt{5}}}{\stackrel{opposite}{1}}\implies csc(\theta )=\sqrt{5}$