\(cos\,x=-\frac{1}{2}\)
\(∴sec\,x=\frac{1}{cos\,x}=\frac{1}{(-\frac{1}{2})}=-2\)
\(sin^2x+cos^2\,x=1\)
\(⇒sin^2+cos^2\,x=1\)
\(⇒sin^2x=1-(-\frac{1}{2})^2\)
\(⇒sin^2x=1-\frac{1}{4}=\frac{3}{4}\)
\(sin^2x=±\frac{±√3}{2}\)
Since x lies in the 3rd quadrant, the value of sin x will be negative.
\(∴sin\,x=-\frac{√3}{2}\)
\(cosecx=\frac{1}{sin\,x}=\frac{1}{-\frac{√3}{2}}=-\frac{2}{√3}\)
\(tan\,x=\frac{sin\,x}{cos \,x}=\frac{-\frac{√3}{2}}{-\frac{1}{2}}=√3\)
\(cot\,x=\frac{1}{tan\,x}=\frac{1}{√3}\)