Question

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

Answer 1

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{aligned}

& ={{(1-\sin \theta +\cos \theta )}^{2}} \\
& ={{(a+b)}^{2}} \\
& ={{a}^{2}}+2ab+{{b}^{2}} \\
\end{aligned}$Now again substituting the values, and keep$2ab$as it is we get,$\begin{aligned}
& ={{a}^{2}}+2ab+{{b}^{2}} \\
& ={{(1-\sin \theta )}^{2}}+{{\cos }^{2}}\theta +2ab \\
\end{aligned}$Simplifying we get,$\begin{aligned}
& =(1-2\sin \theta +{{\sin }^{2}}\theta )+{{\cos }^{2}}\theta +2ab \\
& =1-2\sin \theta +{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2ab \\
\end{aligned}$We know that,${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. So we get,$\begin{aligned}
& =1-2\sin \theta +1+2ab \\
& =1+1-2\sin \theta +2ab \\
& =2-2\sin \theta +2ab \\
\end{aligned}$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.