Question

Using addition formula solve tan 15​

Answer 1

Answer:

2 - $\sqrt{3}$

Step-by-step explanation:

Using the addition formula for tangent

tan(A - B) = $\frac{tanA-tanB}{1+tanAtanB}$ and the exact values

tan45° = 1 , tan60° = $\sqrt{3}$ , then

tan15° = tan(60 - 45)°

tan(60 - 45)°

= $\frac{tan60-tan45}{1+tan60tan45}$

= $\frac{\sqrt{3}-1 }{1+\sqrt{3} }$

Rationalise the denominator by multiplying numerator/ denominator by the conjugate of the denominator.

The conjugate of 1 + $\sqrt{3}$ is 1 - $\sqrt{3}$

= $\frac{(\sqrt{3}-1)(1-\sqrt{3}) }{(1+\sqrt{3})(1-\sqrt{3}) }$ ← expand numerator/denominator using FOIL

= $\frac{\sqrt{3}-3-1+\sqrt{3} }{1-3}$

= $\frac{-4+2\sqrt{3} }{-2}$

= $\frac{-4}{-2}$ + $\frac{2\sqrt{3} }{-2}$

= 2 - $\sqrt{3}$