Answer:
2 - [tex]\sqrt{3}[/tex]
Step-by-step explanation:
Using the addition formula for tangent
tan(A - B) = [tex]\frac{tanA-tanB}{1+tanAtanB}[/tex] and the exact values
tan45° = 1 , tan60° = [tex]\sqrt{3}[/tex] , then
tan15° = tan(60 - 45)°
tan(60 - 45)°
= [tex]\frac{tan60-tan45}{1+tan60tan45}[/tex]
= [tex]\frac{\sqrt{3}-1 }{1+\sqrt{3} }[/tex]
Rationalise the denominator by multiplying numerator/ denominator by the conjugate of the denominator.
The conjugate of 1 + [tex]\sqrt{3}[/tex] is 1 - [tex]\sqrt{3}[/tex]
= [tex]\frac{(\sqrt{3}-1)(1-\sqrt{3}) }{(1+\sqrt{3})(1-\sqrt{3}) }[/tex] ← expand numerator/denominator using FOIL
= [tex]\frac{\sqrt{3}-3-1+\sqrt{3} }{1-3}[/tex]
= [tex]\frac{-4+2\sqrt{3} }{-2}[/tex]
= [tex]\frac{-4}{-2}[/tex] + [tex]\frac{2\sqrt{3} }{-2}[/tex]
= 2 - [tex]\sqrt{3}[/tex]
