[tex]\bf tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\
-------------------------------\\\\
tan(75^o)\implies tan(45^o+30^o)=\cfrac{tan(45^o)+tan(30^o)}{1-tan(45^o)tan(30^o)}
\\\\\\
tan(45^o+30^o)=\cfrac{\frac{sin(45^o)}{cos(45^o)}+\frac{sin(30^o)}{cos(30^o)}}{1-\frac{sin(45^o)}{cos(45^o)}\cdot \frac{sin(30^o)}{cos(30^o)}}[/tex]
[tex]\bf tan(45^o+30^o)=\cfrac{\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}+\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}}{1-\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}\cdot \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}}
\implies
tan(45^o+30^o)=\cfrac{1+\frac{1}{\sqrt{3}}}{1-1\cdot \frac{1}{\sqrt{3}}}[/tex]
[tex]\bf tan(45^o+30^o)=\cfrac{\frac{\sqrt{3}+1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}\implies
tan(45^o+30^o)=\cfrac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}
\\\\\\
tan(45^o+30^o)=\cfrac{\sqrt{3}+1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}-1}\implies
tan(45^o+30^o)=\cfrac{\sqrt{3}+1}{\sqrt{3}-1}[/tex]
so... let's rationalize the denominator.. now, the denominator is √(3) - 1, so, we'll use her conjugate, √(3) + 1, and multiply top and bottom by it, so we end up with a "difference of squares" at the bottom, so, let's do so.
[tex]\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\[/tex]
[tex]\bf \cfrac{\sqrt{3}+1}{\sqrt{3}-1}\cdot \cfrac{\sqrt{3}+1}{\sqrt{3}+1}\implies \cfrac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\implies \cfrac{(\sqrt{3}+1)^2}{(\sqrt{3})^2-(1)^2}
\\\\\\
\cfrac{(\sqrt{3}+1)^2}{3-1}\implies \cfrac{(\sqrt{3})^2+2\sqrt{3}+1^2}{2}\implies \cfrac{3+2\sqrt{3}+1}{2}
\\\\\\
\cfrac{4+2\sqrt{3}}{2}\implies \cfrac{\underline{2}(2+\sqrt{3})}{\underline{2}}\implies 2+\sqrt{3}[/tex]
