Answer: (b) \(\frac{5\pi}6\)
Given that \(cos^{-1}(cos\frac{7\pi}6)\)
We know that \(cos^{-1}(cos\,x)=x\,if\,x\in[0,\pi],\) which is the principle value branch
\(\therefore cos^{-1}(cos\frac{7\pi}6)=cos^{-1}[cos(2\pi-\frac{5\pi}6)]\)
\(cos(2\pi-\frac{5\pi}6)=-cos\,\frac{5\pi}6\)
\(=cos^{-1}(cos(2\pi-\frac{5\pi}6))=cos^{-1}(-cos\frac{5\pi}6)\)
\(=\pi-cos^{-1}(cos\frac{5\pi}6)\)
\(=\pi-cos^{-1}(cos(\pi-\frac{\pi}6))\)
\(=\pi-cos^{-1}(-cos\frac{5\pi}6)\)
\(=\pi-(\pi-cos^{-1}cos\frac{5\pi}6)\)
\(=cos^{-1}(cos(\frac{5\pi}6))\)
\(=\frac{5\pi}6\)
