Find cos y and tan y if csc y = -√6/2 and cot y >0.

Note that according to the definition of [tex]\csc y[/tex] it is true that both sine and cosine are negative, once [tex]\csc y = \dfrac{-\sqrt{6} }{2}[/tex] . Because [tex]\cot y > 0[/tex], this conclusion is true. We basically have

[tex]\boxed{(-a)(1/-b)=a/b \text{ such that } a,b\in\mathbb{R}_{\geq 0}}[/tex]

Sure it is true [tex]\forall y\in\mathbb{R}[/tex] but perhaps this way is better to understand.

In order to find sine, we can use the definition and manipulate the rational expression.

[tex]\csc y = \dfrac{-\sqrt{6} }{2} = \dfrac{-\sqrt{6} / -\sqrt{6} }{2/-\sqrt{6} } = \dfrac{1 }{-\dfrac{2}{\sqrt{6} } }[/tex]

Therefore,

[tex]\sin y =-\dfrac{2}{\sqrt{6} }[/tex]

Here I just divided numerator and denominator by [tex]-\sqrt{6}[/tex].

Now, to find cosine we can use the identity

[tex]\boxed{\sin^2y +\cos ^2y =1}[/tex]

Thus,

[tex]\left(-\dfrac{2}{\sqrt{6} }\right)^2 + \cos ^2y =1 \implies \dfrac{4}{6 } +\cos ^2y =1[/tex]

[tex]\implies \cos ^2y =1 - \dfrac{4}{6 } \implies \cos ^2y =\dfrac{1}{3 } \implies \cos y = \pm \dfrac{\sqrt{1} }{\sqrt{3} } = \pm \dfrac{\sqrt{1} \sqrt{3} }{3} = \pm \dfrac{\sqrt{3} }{3}[/tex]

[tex]\cos y = \pm\dfrac{\sqrt{3} }{3}[/tex]

Once we have [tex]\cot y > 0[/tex], we just consider

[tex]\cos y = -\dfrac{\sqrt{3} }{3}[/tex]

FInally, for tangent, just consider

[tex]\boxed{\tan y := \dfrac{\sin y}{\cos y}}[/tex]

thus,

[tex]\tan y = \dfrac{\sin y}{\cos y} = \dfrac{-\frac{2}{\sqrt{6} }}{-\frac{\sqrt{3} }{3}} = \dfrac{6}{\sqrt{18} } =\dfrac{6}{3\sqrt{2} } =\dfrac{2}{\sqrt{2} } = \sqrt{2}[/tex]