Answer
The required solution is $\frac{\tan \theta +\tan \phi }{1-\tan \theta \tan \phi }$
Work Step by Step
We know the identity $\tan \left( \theta +\phi \right)$ is the sum of the tangent of the first angle and the tangent of second angle divided by 1 minus the product of both angles. The first step is to convert the tan in the form of sin and cos. As per the quotient identity of trigonometry $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$. Therefore, the expression can be simplified as:
$\tan \left( \theta +\phi \right)=\frac{\sin \left( \theta +\phi \right)}{\cos \left( \theta +\phi \right)}$
Now, using the sum and difference identity of sin and cos and then multiplying both the numerator and denominator by $\frac{1}{\cos \theta \cos \phi }$.
$\begin{align}
& \tan \left( \theta +\phi \right)=\frac{\sin \left( \theta +\phi \right)}{\cos \left( \theta +\phi \right)} \\
& =\frac{\sin \theta \cos \phi +\cos \theta \sin \phi }{\cos \theta \cos \phi -\sin \theta \sin \phi } \\
& =\frac{\sin \theta \cos \phi +\cos \theta \sin \phi }{\cos \theta \cos \phi -\sin \theta \sin \phi }\times \frac{\frac{1}{\cos \theta \cos \phi }}{\frac{1}{\cos \theta \cos \phi }}
\end{align}$
Thus, the expression can be further simplified as:
$\begin{align}
& \frac{\sin \theta \cos \phi +\cos \theta \sin \phi }{\cos \theta \cos \phi -\sin \theta \sin \phi }\times \frac{\frac{1}{\cos \theta \cos \phi }}{\frac{1}{\cos \theta \cos \phi }}=\frac{\frac{\sin \theta \cos \phi +\cos \theta \sin \phi }{\cos \theta \cos \phi }}{\frac{cos\theta \cos \phi -\sin \theta \sin \phi }{\cos \theta \cos \phi }} \\
& =\frac{\frac{\sin \theta \cos \phi }{\cos \theta \cos \phi }+\frac{\cos \theta \sin \phi }{\cos \theta \cos \phi }}{\frac{cos\theta \cos \phi }{\cos \theta \cos \phi }-\frac{\sin \theta \sin \phi }{\cos \theta \cos \phi }} \\
& =\frac{\frac{\sin \theta }{\cos \theta }.1+1.\frac{\sin \phi }{\cos \phi }}{1-\frac{\sin \theta }{\cos \theta }.\frac{\sin \phi }{\cos \phi }} \\
& =\frac{\tan \theta +\tan \phi }{1-\tan \theta \tan \phi }
\end{align}$
