Question

Pre Calculus!! Is this answer right?!?!? help!!

Answer 1

so hmmm we know cos(θ) < 0, which is another way of saying the cosine is negative, we also know the cosecant is negative, so its reciprocal is also negative, namely the sine, so sine and cosine are both negative, the only occurs in the III Quadrant. Let's also recall that the hypotenuse is always positive.

`\csc( \theta )=-\cfrac{\stackrel{hypotenuse}{√(5)}}{\underset{opposite}{2}} \implies \csc( \theta )=\cfrac{\stackrel{hypotenuse}{√(5)}}{\underset{opposite}{-2}}~\hfill~\textit{let's find the \underline{adjacent}} \\\\\\ \begin{array}{llll} \textit{using the pythagorean theorem} \\\\ a^2+o^2=c^2\implies a=√(c^2 - o^2) \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{√(5)}\\ a=adjacent\\ o=\stackrel{opposite}{-2} \end{cases}`

`a=\pm \sqrt{ (√(5))^2 - (-2)^2}\implies a=\pm √( 5 - 4 ) \implies a=\pm 1\implies \stackrel{ \textit{III Quadrant} }{a=-1} \\\\[-0.35em] ~\dotfill\\\\ \sin(\theta )=\cfrac{\stackrel{opposite}{-2}}{\underset{hypotenuse}{√(5)}}\implies \sin(\theta )=-\cfrac{2√(5)}{5} \\\\\\ \cos(\theta )=\cfrac{\stackrel{adjacent}{-1}}{\underset{hypotenuse}{√(5)}}\implies \cos(\theta )=-\cfrac{√(5)}{5}`

`\tan(\theta )=\cfrac{\stackrel{opposite}{-2}}{\underset{adjacent}{-1}}\implies \tan(\theta )=2 \\\\\\ \sec(\theta )=\cfrac{\stackrel{hypotenuse}{√(5)}}{\underset{adjacent}{-1}}\implies \sec(\theta )=-√(5) \\\\\\ \cot(\theta )=\cfrac{\stackrel{adjacent}{-1}}{\underset{opposite}{-2}}\implies \cot(\theta )=\cfrac{1}{2}`