## Precalculus (6th Edition) Blitzer

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.