Is tan((7pi)/6)=(sqrt(3))/3 ?

The answer to whether tan((7pi)/6)=(sqrt(3))/3 is True

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prove tan((7pi)/6)=(sqrt(3))/3

prove

tan (\frac{7 π}{6}) = \frac{3}{3}

True

Solution steps

tan (\frac{7 π}{6}) = \frac{3}{3}

Manipulating left side

tan (\frac{7 π}{6})

Simplify

tan (\frac{7 π}{6}) : \frac{3}{3}

tan (\frac{7 π}{6})

tan (\frac{7 π}{6}) = tan (\frac{π}{6})

tan (\frac{7 π}{6})

Rewrite

\frac{7 π}{6}

π + \frac{π}{6}

= tan (π + \frac{π}{6})

Apply the periodicity of

tan

tan (x + π) = tan (x)

tan (π + \frac{π}{6}) = tan (\frac{π}{6})

= tan (\frac{π}{6})

= tan (\frac{π}{6})

Use the following trivial identity:

tan (\frac{π}{6}) = \frac{3}{3}

tan (\frac{π}{6})

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= \frac{3}{3}

= \frac{3}{3}

= \frac{3}{3}

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 - cos ^{2} ( x )}{tan ( x )} = sin (x) cos (x)

- tan (θ) - \frac{cos ( θ )}{sin ( θ ) - 1} = sec (θ)

cot (θ) \cdot sec (θ) = csc (θ)

csc^{2} (x) + cot^{2} (x) = 1 + 2 cot^{2} (x)

csc (x) - \frac{cos ^{2} ( x )}{sin ( x )} = sin (x)

Is tan((7pi)/6)=(sqrt(3))/3 ?
The answer to whether tan((7pi)/6)=(sqrt(3))/3 is True