\(y=\frac{1}{\sqrt{sin^2x}}=\left(sin^2x\right)^{-\frac{1}{2}}\)
\(\Rightarrow y'=-\frac{1}{2}\left(sin^2x\right)^{-\frac{3}{2}}.2sinx.cosx=\frac{-sinx.cosx}{\sqrt{sin^6x}}=\frac{-sinx.cosx}{\sqrt{sin^2x.sin^4x}}=\frac{-sinx.cox}{\left|sinx\right|.sin^2x}=\frac{-cosx}{\left|sinx\right|.sinx}\)