Answer
The required solution is False.
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 the second angle divided by 1 minus the products of both angles.
$\tan \left( x+y \right)=\frac{\tan x+\tan y}{1-\tan x\tan y}$
Then, replacing the values in the above equation, we get:
$\begin{align}
& \tan {{75}^{\circ }}=\tan \left( {{30}^{\circ }}+{{45}^{\circ }} \right)=\frac{\tan {{30}^{\circ }}+\tan {{45}^{\circ }}}{1-\tan {{30}^{\circ }}\tan {{45}^{\circ }}} \\
& =\frac{\frac{1}{\sqrt{3}}+1}{1-\frac{1}{\sqrt{3}}.1} \\
& =\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}
\end{align}$
$\begin{align}
& =\frac{1+\frac{1}{1.732}}{1-\frac{1}{1.732}} \\
& =\frac{1+0.577}{1-0.577} \\
& =\frac{1.577}{0.423} \\
& =3.728 \\
\end{align}$
$\begin{align}
& \tan {{30}^{o}}+\tan {{45}^{o}}=\frac{1}{\sqrt{3}}+1 \\
& \text{ = 0}\text{.577+1} \\
& \text{ = 1}\text{.577} \\
\end{align}$
Thus, $\tan {{75}^{o}}\ne \tan {{30}^{o}}+\tan {{45}^{o}}$
Hence, the given statement is false.
