Question

The exact value of tan 5 π/12

Answer 1

Answer:

C

Step-by-step explanation:

Using the addition formula for tangent

tan(x + y) = $\frac{tanx+tany}{1-tanxtany}$ and the exact values

tan($\frac{\pi }{6}$ ) = $\frac{1}{\sqrt{3} }$ and tan($\frac{\pi }{4}$ ) = 1

tan($\frac{5\pi }{12}$ ) = tan($\frac{\pi }{6}$ + $\frac{\pi }{4}$ )

= $\frac{tan(\frac{\pi }{6})+tan(\frac{\pi }{4}) }{1-tan(\frac{\pi }{6})tan(\frac{\pi }{4}) }$

= $\frac{\frac{1}{\sqrt{3} }+1 }{1-\frac{1}{\sqrt{3} }.1 }$  ( multiply numerator/ denominator by $\sqrt{3}$ )

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

rationalise denominator by multiplying numerator/denominator by ($\sqrt{3}$ + 1)

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

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

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

= $\sqrt{3}$ + 2 → C