What is the value of -tan^2((4pi)/3) ?

The value of -tan^2((4pi)/3) is -3

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-tan^2((4pi)/3)

- tan^{2} (\frac{4 π}{3})

- 3

Solution steps

- tan^{2} (\frac{4 π}{3})

Rewrite using trig identities:

tan (\frac{4 π}{3}) = 3

tan (\frac{4 π}{3})

tan (\frac{4 π}{3}) = tan (\frac{π}{3})

tan (\frac{4 π}{3})

Rewrite

\frac{4 π}{3}

π + \frac{π}{3}

= tan (π + \frac{π}{3})

Apply the periodicity of

tan

tan (x + π) = tan (x)

tan (π + \frac{π}{3}) = tan (\frac{π}{3})

= tan (\frac{π}{3})

= tan (\frac{π}{3})

Use the following trivial identity:

tan (\frac{π}{3}) = 3

tan (\frac{π}{3})

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}

= 3

= 3

= - (3)^{2}

Simplify

- (3)^{2} : - 3

- (3)^{2}

Simplify

(3)^{2} : 3

(3)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

= - 3

= - 3

2 sin (4 8^{\circ}) cos (4 8^{\circ})

csc (- 3)

sin (62. 5^{\circ})

cos (5 \frac{π}{6})

sec (- 4)