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Answer:
[tex]\displaystyle \tan\frac{5\pi}{12}=2+\sqrt{3}[/tex]
Step-by-step explanation:
Tangent Half Angle
Given an angle θ, then:
[tex]\displaystyle \tan {\frac {\theta }{2}}=\frac {\sin \theta }{1+\cos \theta}[/tex]
We are required to find:
[tex]\tan\frac{5\pi}{12}[/tex]
But it cannot be found in tables of main angles. We can use the angle
[tex]\theta = \frac{5\pi}{6}[/tex]
And use the formula above to find the required operation. Hence:
[tex]\displaystyle \tan\frac{5\pi}{12}=\tan\frac{\frac{5\pi}{6}}{2}[/tex]
[tex]\displaystyle \tan\frac{5\pi}{12}=\frac {\sin \frac{5\pi}{6} }{1+\cos \frac{5\pi}{6}}[/tex]
[tex]\displaystyle \tan\frac{5\pi}{12}=\frac {\frac{1}{2} }{1-\frac{\sqrt{3}}{2}}[/tex]
Operating:
[tex]\displaystyle \tan\frac{5\pi}{12}=\frac {\frac{1}{2} }{\frac{2-\sqrt{3}}{2}}[/tex]
Simplifying:
[tex]\displaystyle \tan\frac{5\pi}{12}=\frac{1}{2-\sqrt{3}}[/tex]
Rationalizing:
[tex]\displaystyle \tan\frac{5\pi}{12}=\frac{1}{2-\sqrt{3}}\cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}[/tex]
[tex]\displaystyle \tan\frac{5\pi}{12}=\frac{2+\sqrt{3}}{2^2-\sqrt{3}^2}[/tex]
[tex]\displaystyle \tan\frac{5\pi}{12}=\frac{2+\sqrt{3}}{4-3}[/tex]
Finally:
[tex]\boxed{\displaystyle \tan\frac{5\pi}{12}=2+\sqrt{3}}[/tex]