Question

Find the exact value of tan75

This is a question from sum and differences of trigs so it must include that. I've tried multiple things and nothing has given me one of the options for answers.

Answer 1

`\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{(sin(45^o))/(cos(45^o))+(sin(30^o))/(cos(30^o))}{1-(sin(45^o))/(cos(45^o))\cdot (sin(30^o))/(cos(30^o))}`


`\bf tan(45^o+30^o)=\cfrac{((√(2))/(2))/((√(2))/(2))+((1)/(2))/((√(3))/(2))}{1-((√(2))/(2))/((√(2))/(2))\cdot ((1)/(2))/((√(3))/(2))} \implies tan(45^o+30^o)=\cfrac{1+(1)/(√(3))}{1-1\cdot (1)/(√(3))}`


`\bf tan(45^o+30^o)=\cfrac{(√(3)+1)/(√(3))}{1-(1)/(√(3))}\implies tan(45^o+30^o)=\cfrac{(√(3)+1)/(√(3))}{(√(3)-1)/(√(3))} \\\\\\ tan(45^o+30^o)=\cfrac{√(3)+1}{√(3)}\cdot \cfrac{√(3)}{√(3)-1}\implies tan(45^o+30^o)=\cfrac{√(3)+1}{√(3)-1}`


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.


`\bf \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -------------------------------\\\\`


`\bf \cfrac{√(3)+1}{√(3)-1}\cdot \cfrac{√(3)+1}{√(3)+1}\implies \cfrac{(√(3)+1)(√(3)+1)}{(√(3)-1)(√(3)+1)}\implies \cfrac{(√(3)+1)^2}{(√(3))^2-(1)^2} \\\\\\ \cfrac{(√(3)+1)^2}{3-1}\implies \cfrac{(√(3))^2+2√(3)+1^2}{2}\implies \cfrac{3+2√(3)+1}{2} \\\\\\ \cfrac{4+2√(3)}{2}\implies \cfrac{\underline{2}(2+√(3))}{\underline{2}}\implies 2+√(3)`