Question

Evaluate: $\tan\left(\frac{5\pi}{6}\right) + \cos\left(\frac{2\pi}{3}\right)$. \\ $-\frac{5\sqrt{3}}{6}$ \\ $-\frac{4\sqrt{3}}{3}$ \\ $-\frac{2\sqrt{3}+3}{6}$ \\ $-\frac{2\sqrt{3}-3}{6}$ \\ $-\frac{\sqrt{3}+1}{2}$ \\ None of the above.

          Evaluate: $\tan\left(\frac{5\pi}{6}\right) + \cos\left(\frac{2\pi}{3}\right)$. \\
$-\frac{5\sqrt{3}}{6}$ \\
$-\frac{4\sqrt{3}}{3}$ \\
$-\frac{2\sqrt{3}+3}{6}$ \\
$-\frac{2\sqrt{3}-3}{6}$ \\
$-\frac{\sqrt{3}+1}{2}$ \\
None of the above.

Evaluate: tan((5π)/(6)) + cos((2π)/(3)).

-(5√(3))/(6)

-(4√(3))/(3)

-(2√(3)+3)/(6)

-(2√(3)-3)/(6)

-(√(3)+1)/(2)

None of the above.

Added by Anna S.

Precalculus with Limits

Ron Larson 2nd Edition

Instant Answer

Solved by Expert Faizanullah Kazmi

05/24/2021

Step 1

Step 1: tan($$\frac{5\pi}{6}$$) = -$$\frac{1}{\sqrt{3}}$$ Show more…

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Evaluate: tan($$\frac{5\pi}{6}$$) + cos($$\frac{2\pi}{3}$$). - $$\frac{5\sqrt{3}}{6}$$ - $$\frac{4\sqrt{3}}{3}$$ - $$\frac{2\sqrt{3}+3}{6}$$ - $$\frac{2\sqrt{3}-3}{6}$$ - $$\frac{\sqrt{3}+1}{2}$$ None of the above.