Answer:
sin x=[tex]-\frac{\sqrt5}{3}[/tex]
[tex] csc(x)=-\frac{3}{\sqrt5}[/tex]
[tex]tan x=\frac{\sqrt5}{2}[/tex]
[tex] cot x=\frac{2}{\sqrt5}[/tex]
Step-by-step explanation:
We are given that
[tex] sec x=-\frac{3}{2}[/tex], cos (x)=[tex]-\frac{2}{3}[/tex]
tan x >0
We have to find sinx , csc (x), tan (x), cot (x).
We know that [tex] sin x=\sqrt{1-cos^2x}[/tex]
Using the formula
Then, we get
[tex] sin x=\sqrt{1-(\frac{-2}{3})^2}=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{9-4}{9}}=\frac{\sqrt5}{3}[/tex]
We are given that tan x >0 and cos x(x) <0
It means angle x lies in III quadrant.
In III quadrant , sin x and cos x are both negative and tan x is positive.
Therefore , sin x=[tex]-\frac{\sqrt5}{3}[/tex]
[tex] csc (x)=\frac{1}{sinx}[/tex]
[tex] csc(x)=\frac{1}{\frac{-\sqrt5}{3}}=-\frac{3}{\sqrt5}[/tex]
[tex] csc(x)=-\frac{3}{\sqrt5}[/tex]
[tex]\tan x=\frac{sinx}{cosx}[/tex]
[tex]tan x=\frac{-\frac{\sqrt5}{3}}{-\frac{2}{3}}=\frac{\sqrt5}{2}[/tex]
[tex]tan x=\frac{\sqrt5}{2}[/tex]
[tex] cot x=\frac{cos x}{sinx}[/tex]
[tex]cot x=\frac{-\frac{2}{3}}{-\frac{\sqrt5}{3}}=\frac{2}{\sqrt5}[/tex]
[tex] cot x=\frac{2}{\sqrt5}[/tex]
