\(sin\,x=\frac{3}{2}\)
\(cosec\,x=\frac{1}{sin\,x}=\frac{1}{(\frac{3}{5})}=\frac{5}{3}\)
\(sin^2x+cos^2\,x=1\)
\(⇒cos^2x=1-sin^2x\)
\(⇒cos^2x=1-(\frac{3}{5})^2\)
\(⇒cos^2x=1-\frac{9}{25}\)
\(⇒cos^2\,x=\frac{16}{25}\)
\(⇒cos^2\,x=±\frac{4}{5}\)
Since x lies in the 2nd quadrant, the value of cos x will be negative
\(∴cos\,x=-\frac{4}{5}\)
\(sec\,x=\frac{1}{cos\,\,x}=\frac{1}{(-\frac{4}{5})}=-\frac{5}{4}\)
\(tan\,x=\frac{sin\,x}{cos \,x}=\frac{(\frac{3}{5})}{(-\frac{4}{5})}=\frac{3}{4}\)
\(cot\,x=\frac{1}{tan\,x}=\frac{4}{3}\)