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

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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.
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