What is the value of (tan(300))/(sin(120)) ?

The value of (tan(300))/(sin(120)) is -2

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(tan(300))/(sin(120))

\frac{tan ( 30 0 ^{\circ} )}{sin ( 12 0 ^{\circ} )}

- 2

Solution steps

\frac{tan ( 30 0 ^{\circ} )}{sin ( 12 0 ^{\circ} )}

tan (30 0^{\circ}) = tan (12 0^{\circ})

tan (30 0^{\circ})

Rewrite

30 0^{\circ}

18 0^{\circ} + 12 0^{\circ}

= tan (18 0^{\circ} + 12 0^{\circ})

Apply the periodicity of

tan

tan (x + 18 0^{\circ}) = tan (x)

tan (18 0^{\circ} + 12 0^{\circ}) = tan (12 0^{\circ})

= tan (12 0^{\circ})

= \frac{tan ( 12 0 ^{\circ} )}{sin ( 12 0 ^{\circ} )}

Rewrite using trig identities:

tan (12 0^{\circ}) = \frac{sin ( 12 0 ^{\circ} )}{cos ( 12 0 ^{\circ} )}

tan (12 0^{\circ})

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( 12 0 ^{\circ} )}{cos ( 12 0 ^{\circ} )}

= \frac{\frac{s i n ( 12 0 ^{\circ} )}{c o s ( 12 0 ^{\circ} )}}{sin ( 12 0 ^{\circ} )}

Simplify

= \frac{1}{cos ( 12 0 ^{\circ} )}

Use the following trivial identity:

cos (12 0^{\circ}) = - \frac{1}{2}

cos (12 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{1}{2}

= \frac{1}{- \frac{1}{2}}

Simplify

= - 2

cos (\frac{8 π}{7})

15 cos (5 0^{\circ})

arctan (\frac{2}{- 4})

4 + 3 cos (- \frac{π}{3})

arctan (0.64)

What is the value of (tan(300))/(sin(120)) ?
The value of (tan(300))/(sin(120)) is -2