Hint: We have to prove that ${{(1-\sin \theta +\cos \theta )}^{2}}=2(1+\cos \theta )(1-\sin \theta )$, for that you should prove LHS$=$RHS. Assume $(1-\sin \theta )=a$ and $\cos \theta =b$, and simplify. Try it, you will get the answer.
Now we are given ${{(1-\sin \theta +\cos \theta )}^{2}}=2(1+\cos \theta )(1-\sin \theta )$. for that we have to prove LHS$=$RHS, So first let us consider, $(1-\sin \theta )=a$ and $\cos \theta =b$. So substituting the values in the LHS we get, $\begin{align}
& ={{(1-\sin \theta +\cos \theta )}^{2}} \\ & ={{(a+b)}^{2}} \\ & ={{a}^{2}}+2ab+{{b}^{2}} \\ \end{align}$ Now again substituting the values, and keep $2ab$ as it is we get, $\begin{align} & ={{a}^{2}}+2ab+{{b}^{2}} \\ & ={{(1-\sin \theta )}^{2}}+{{\cos }^{2}}\theta +2ab \\ \end{align}$ Simplifying we get, $\begin{align} & =(1-2\sin \theta +{{\sin }^{2}}\theta )+{{\cos }^{2}}\theta +2ab \\ & =1-2\sin \theta +{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2ab \\ \end{align}$ We know that, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. So we get, $\begin{align} & =1-2\sin \theta +1+2ab \\ & =1+1-2\sin \theta +2ab \\ & =2-2\sin \theta +2ab \\ \end{align}$ Taking $2$common we get, $=2(1-\sin \theta +ab)$ We know that, $(1-\sin \theta )=a$ so substituting we get, $=2(a+ab)=2a(1+b)$ Now let us take RHS$=2(1+\cos \theta )(1-\sin \theta )$. Substituting the values $(1-\sin \theta )=a$ and $\cos \theta =b$ we get, RHS$=2a(1+b)$ So we can see that LHS$=$RHS.
Note: Read the question carefully. Also, do not make silly mistakes. Assumption you take should be right, as we took here $(1-\sin \theta )=a$ and $\cos \theta =b$. Do not make any mistake. Take utmost care that no confusion occurs.
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