Answer
The result of $\cos \left( \alpha +\beta \right)\cos \beta +\sin \left( \alpha +\beta \right)\sin \beta $ is $\cos \alpha $.
Work Step by Step
Let us consider the given expression,
$\cos \left( \alpha +\beta \right)\cos \beta +\sin \left( \alpha +\beta \right)\sin \beta $
By using the trigonometric identities,
$\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta $
$\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta $
Now, the above expression can be further simplified as:
$\begin{align}
& \cos \left( \alpha +\beta \right)\cos \beta +\sin \left( \alpha +\beta \right)\sin \beta =\left\{ \left( \cos \alpha \cos \beta -\sin \alpha \sin \beta \right)\cos \beta +\left( \sin \alpha \cos \beta +\cos \alpha \sin \beta \right)\sin \beta \right\} \\
& =\left\{ \cos \alpha {{\cos }^{2}}\beta -\sin \alpha \sin \beta \cos \beta +\sin \alpha \cos \beta \sin \beta +\cos \alpha {{\sin }^{2}}\beta \right\} \\
& =\cos \alpha {{\cos }^{2}}\beta +\cos \alpha {{\sin }^{2}}\beta \\
& =\cos \alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right)
\end{align}$
by using the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
$\begin{align}
& \cos \alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right)=\cos \alpha \left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right) \\
& =\cos \alpha \left( 1 \right) \\
& =\cos \alpha
\end{align}$
Thus, $\cos \left( \alpha +\beta \right)\cos \beta +\sin \left( \alpha +\beta \right)\sin \beta $ can be simplified as $\cos \alpha $.
Hence, the result of $\cos \left( \alpha +\beta \right)\cos \beta +\sin \left( \alpha +\beta \right)\sin \beta $ is $\cos \alpha $.
