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

Question

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

Vedantu Content Team · Accepted Answer

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

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

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

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