What is the value of tan^2((2pi)/3) ?

The value of tan^2((2pi)/3) is 3

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tan^2((2pi)/3)

tan^{2} (\frac{2 π}{3})

3

Solution steps

tan^{2} (\frac{2 π}{3})

Rewrite using trig identities:

sec^{2} (\frac{2 π}{3}) - 1

tan^{2} (\frac{2 π}{3})

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= sec^{2} (\frac{2 π}{3}) - 1

= sec^{2} (\frac{2 π}{3}) - 1

Use the following trivial identity:

sec (\frac{2 π}{3}) = - 2

sec (\frac{2 π}{3})

sec (x)

periodicity table with

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

= - 2

= (- 2)^{2} - 1

Simplify

(- 2)^{2} - 1 : 3

(- 2)^{2} - 1

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 2^{2} - 1

2^{2} = 4

= 4 - 1

Subtract the numbers:

4 - 1 = 3

= 3

= 3

2 \frac{sin ( 0 )}{0}

4 sec (6 0^{\circ})

arctan (\frac{24}{12})

sin^{2} (9 0^{\circ}) - cos^{2} (9 0^{\circ})

\frac{65 \cdot 9.8}{cos ( 3 8 ^{\circ} )}