Question

Tan (pi/5) - tan (pi/30) / 1+tan (pi/5)tan (pi/30)

Answer 1

Find the value using the definition of tangent.tan(+)=opposite/adjacent tan(+)= opposite/adjacentSubstitute the values into the definition.tan(+)=0/1 tan(+)=0/1Divide 0/0 by 1/1 to get 0/0.Answer is 0

Answer 2

$\bf \textit{Sum and Difference Identities} \\ \quad \\ sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}}) \\ \quad \\ sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} - {{ \beta}})= cos({{ \alpha}})cos({{ \beta}}) + sin({{ \alpha}})sin({{ \beta}}) \\ \quad$

$\bf tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\qquad tan({{ \alpha}} - {{ \beta}}) = \cfrac{tan({{ \alpha}})- tan({{ \beta}})}{1+ tan({{ \alpha}})tan({{ \beta}})}\\\\ -------------------------------$

$\bf \cfrac{tan\left( \frac{\pi }{5} \right)-tan\left( \frac{\pi }{30} \right)}{1+tan\left( \frac{\pi }{5} \right)tan\left( \frac{\pi }{30} \right)}\implies tan\left( \frac{\pi }{5} -\frac{\pi }{30} \right)\implies tan\left( \frac{6\pi -\pi }{30} \right)$

$\bf tan\left( \frac{5\pi }{30} \right)\implies tan\left( \frac{\pi }{6} \right)\implies \cfrac{sin\left( \frac{\pi }{6} \right)}{cos\left( \frac{\pi }{6} \right)}\implies \cfrac{\quad \frac{1}{2}\quad }{\frac{\sqrt{3}}{2}} \\\\\\ \cfrac{1}{2}\cdot \cfrac{2}{\sqrt{3}}\implies \cfrac{1}{\sqrt{3}}\implies \cfrac{\sqrt{3}}{3}$