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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.

${{(1-\sin \theta +\cos \theta )}^{2}}=2(1+\cos \theta )(1-\sin \theta )$

Hence proved.

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.

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.

${{(1-\sin \theta +\cos \theta )}^{2}}=2(1+\cos \theta )(1-\sin \theta )$

Hence proved.

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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