Question

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

Answer 1

Answer:

$\cos y = -\dfrac{\sqrt{3} }{3}$

$\tan y = \sqrt{2}$

Step-by-step explanation:

Recall that

$\boxed{\csc y := \dfrac{1}{\sin y}}$

$\boxed{\cot y := \dfrac{\cos y}{\sin y}}$

We know that

$\csc y = \dfrac{-\sqrt{6} }{2}$

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

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

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

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

$\csc y = \dfrac{-\sqrt{6} }{2} = \dfrac{-\sqrt{6} / -\sqrt{6} }{2/-\sqrt{6} } = \dfrac{1 }{-\dfrac{2}{\sqrt{6} } }$

Therefore,

$\sin y =-\dfrac{2}{\sqrt{6} }$

Here I just divided numerator and denominator by $-\sqrt{6}$.

Now, to find cosine we can use the identity

$\boxed{\sin^2y +\cos ^2y =1}$

Thus,

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

$\implies \cos ^2y =1 - \dfrac{4}{6 } \implies \cos ^2y =\dfrac{1}{3 } \implies \cos y = \pm \dfrac{\sqrt{1} }{\sqrt{3} } = \pm \dfrac{\sqrt{1} \sqrt{3} }{3} = \pm \dfrac{\sqrt{3} }{3}$

$\cos y = \pm\dfrac{\sqrt{3} }{3}$

Once we have $\cot y > 0$, we just consider

$\cos y = -\dfrac{\sqrt{3} }{3}$

FInally, for tangent, just consider

$\boxed{\tan y := \dfrac{\sin y}{\cos y}}$

thus,

$\tan y = \dfrac{\sin y}{\cos y} = \dfrac{-\frac{2}{\sqrt{6} }}{-\frac{\sqrt{3} }{3}} = \dfrac{6}{\sqrt{18} } =\dfrac{6}{3\sqrt{2} } =\dfrac{2}{\sqrt{2} } = \sqrt{2}$