Answer
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Hint: We will be using the concept of trigonometric identities to solve the problem. We will first square the equations \[\cos A+\sin B=m\ and\ \sin A+\cos B=n\] and then add them to further simplify the solution.
Complete step-by-step answer:
Now, we have been given that,
\[\begin{align}
& \cos A+\sin B=m............\left( 1 \right) \\
& \sin A+\cos B=n............\left( 2 \right) \\
\end{align}\]
We have to prove that,
$2\sin \left( A+B \right)={{m}^{2}}+{{n}^{2}}-2$
Now, we will first square (1) and (2). So, we have,
\[\begin{align}
& {{\left( \cos A+\sin B \right)}^{2}}={{m}^{2}} \\
& {{\left( \sin A+\cos B \right)}^{2}}={{n}^{2}} \\
\end{align}\]
Now, we know the algebraic identity that,
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
So, we have,
\[\begin{align}
& {{\cos }^{2}}A+{{\sin }^{2}}B+2\cos A\sin B={{m}^{2}}............\left( 3 \right) \\
& {{\sin }^{2}}A+{{\cos }^{2}}B+2\sin A\cos B={{n}^{2}}............\left( 4 \right) \\
\end{align}\]
Now, we will add (3) and (4). So, we have,
$\left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)+\left( {{\sin }^{2}}B+{{\cos }^{2}}B \right)+2\cos
A\sin B+2\sin A\cos B={{m}^{2}}+{{n}^{2}}$
Now, we know the trigonometric identity that,
$\begin{align}
& {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\
& {{\sin }^{2}}B+{{\cos }^{2}}B=1 \\
\end{align}$
So, we will use this to further simplify the equation. So, we have,
$1+1+2\sin A\cos B+2\cos A\sin B={{m}^{2}}+{{n}^{2}}$
Now, we will rearrange the terms as,
$2\left( \sin A\cos B+\cos A\sin B \right)={{m}^{2}}+{{n}^{2}}-2$
Now, we know the trigonometric identity that,
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
So, we have by using this that,
$2\sin \left( A+B \right)={{m}^{2}}+{{n}^{2}}-2$
Now, since LHS = RHS.
Hence, proved.
Note: To solve these type of questions it is important to note how we have squared equation (1) and (2) and added those to get the desired equation one must note the hint for this the terms like ${{m}^{2}}\ and\ {{n}^{2}}$ appearing in the equation to be proved.
Complete step-by-step answer:
Now, we have been given that,
\[\begin{align}
& \cos A+\sin B=m............\left( 1 \right) \\
& \sin A+\cos B=n............\left( 2 \right) \\
\end{align}\]
We have to prove that,
$2\sin \left( A+B \right)={{m}^{2}}+{{n}^{2}}-2$
Now, we will first square (1) and (2). So, we have,
\[\begin{align}
& {{\left( \cos A+\sin B \right)}^{2}}={{m}^{2}} \\
& {{\left( \sin A+\cos B \right)}^{2}}={{n}^{2}} \\
\end{align}\]
Now, we know the algebraic identity that,
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
So, we have,
\[\begin{align}
& {{\cos }^{2}}A+{{\sin }^{2}}B+2\cos A\sin B={{m}^{2}}............\left( 3 \right) \\
& {{\sin }^{2}}A+{{\cos }^{2}}B+2\sin A\cos B={{n}^{2}}............\left( 4 \right) \\
\end{align}\]
Now, we will add (3) and (4). So, we have,
$\left( {{\sin }^{2}}A+{{\cos }^{2}}A \right)+\left( {{\sin }^{2}}B+{{\cos }^{2}}B \right)+2\cos
A\sin B+2\sin A\cos B={{m}^{2}}+{{n}^{2}}$
Now, we know the trigonometric identity that,
$\begin{align}
& {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\
& {{\sin }^{2}}B+{{\cos }^{2}}B=1 \\
\end{align}$
So, we will use this to further simplify the equation. So, we have,
$1+1+2\sin A\cos B+2\cos A\sin B={{m}^{2}}+{{n}^{2}}$
Now, we will rearrange the terms as,
$2\left( \sin A\cos B+\cos A\sin B \right)={{m}^{2}}+{{n}^{2}}-2$
Now, we know the trigonometric identity that,
$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$
So, we have by using this that,
$2\sin \left( A+B \right)={{m}^{2}}+{{n}^{2}}-2$
Now, since LHS = RHS.
Hence, proved.
Note: To solve these type of questions it is important to note how we have squared equation (1) and (2) and added those to get the desired equation one must note the hint for this the terms like ${{m}^{2}}\ and\ {{n}^{2}}$ appearing in the equation to be proved.
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