Question

Prove the following, if \[\cos \theta -\sin \theta =1\]:
  \[\cos \theta +\sin \theta =1\]

Answer 1

Hint: In such questions, we can think of squaring both the sides of the equation and then we can try to solve this question.
The important trigonometric identity that would be used in solving this question is as follows
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Complete step-by-step answer:
As mentioned in the question, we have to prove that \[\cos \theta +\sin \theta =1\].
In the given equation, we can square both the sides as mentioned in the hint as follows
\[\begin{align}
  & {{\left( \cos \theta -\sin \theta \right)}^{2}}={{1}^{2}} \\
 & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta -2\cos \theta \sin \theta =1\ \ \ \ \ ...(1) \\
\end{align}\]
Now, on using the identity that is mentioned in the hint, we can proceed as follows
\[\begin{align}
  & 1-2\sin \theta \cos \theta =1 \\
 & 2\sin \theta \cos \theta =0\ \ \ \ \ ...(a) \\
\end{align}\]
Now, using this equation (a), we will prove the required equation as follows
Now, adding \[4\cos \theta \sin \theta \] on both the sides in equation (1), we get the following result
\[\begin{align}
  & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\sin \theta \cos \theta =1+4\sin \theta \cos \theta \\
 & {{\left( \cos \theta +\sin \theta \right)}^{2}}=1+0 \\
 & {{\left( \cos \theta +\sin \theta \right)}^{2}}=1 \\
 & \cos \theta +\sin \theta =1 \\
\end{align}\]
Hence, on taking the square root on both sides in the second last step, we thus proved that
\[\cos \theta +\sin \theta =1\] .

Note: The most important identity for this type of question is $(a - b)^2+4ab$ = $(a + b)^2$ and trigonometric identity that would be used in solving this question is as follows
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]