It can be written as
tan15° = tan (45° – 30°)
Using formula
\( \dfrac{tan45°-tan30°}{1+tan45°+tan 30°}\)
\( tan(x-y)= \dfrac{tanx-tany}{1+tanxtany}\)
Since we know,
Therefore we have,
\( tan15°=\dfrac{1-\dfrac{1}{\sqrt{3}}}{1+1\dfrac{1}{\sqrt{3}}}\)
= \( \dfrac{\dfrac{\sqrt{3}-1}{\sqrt{3}}}{\dfrac{\sqrt{3}+1}{\sqrt{3}}}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1} \)
= \( \dfrac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} \)
Further computing we have,
tan15° = \( \dfrac{3+1-2\sqrt{3}}{(\sqrt{3})^2-(1)^2}\)
= \( \dfrac{4-2\sqrt{3}}{3-1}\)
= \( 2-\sqrt{3}\)
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