If  $\sin x+{\sin }^{2}x=1$ then the value of ${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x-1$ is equal to

From $\sin x+{\sin }^{2}x=1\text{, we get }\sin x={\cos }^{2}x$

 Now, the given expression is equal to

  $\begin{array}{l}{\cos }^{6}x\left({\cos }^{6}x+3{\cos }^{4}x+3{\cos }^{2}x+1\right)-1\\ ={\cos }^{6}x{\left({\cos }^{2}x+1\right)}^3-1\\ ={\sin }^{3}x(\sin x+1{)}^3-1\\ ={\left({\sin }^{2}x+\sin x\right)}^3-1=1-1=0\end{array}$