What is the value of sin(150)+cos(510)+tan(4110)-tan(210) ?

The value of sin(150)+cos(510)+tan(4110)-tan(210) is -(7sqrt(3)-3)/6

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sin(150)+cos(510)+tan(4110)-tan(210)

sin (15 0^{\circ}) + cos (51 0^{\circ}) + tan (411 0^{\circ}) - tan (21 0^{\circ})

- \frac{7 3 - 3}{6}

Solution steps

sin (15 0^{\circ}) + cos (51 0^{\circ}) + tan (411 0^{\circ}) - tan (21 0^{\circ})

cos (51 0^{\circ}) = cos (15 0^{\circ})

= sin (15 0^{\circ}) + cos (15 0^{\circ}) + tan (411 0^{\circ}) - tan (21 0^{\circ})

tan (411 0^{\circ}) = tan (15 0^{\circ})

= sin (15 0^{\circ}) + cos (15 0^{\circ}) + tan (15 0^{\circ}) - tan (21 0^{\circ})

tan (21 0^{\circ}) = tan (3 0^{\circ})

= sin (15 0^{\circ}) + cos (15 0^{\circ}) + tan (15 0^{\circ}) - tan (3 0^{\circ})

Rewrite using trig identities:

tan (15 0^{\circ}) = \frac{sin ( 15 0 ^{\circ} )}{cos ( 15 0 ^{\circ} )}

= sin (15 0^{\circ}) + cos (15 0^{\circ}) + \frac{sin ( 15 0 ^{\circ} )}{cos ( 15 0 ^{\circ} )} - tan (3 0^{\circ})

Simplify

= \frac{sin ( 15 0 ^{\circ} ) cos ( 15 0 ^{\circ} ) + cos ^{2} ( 15 0 ^{\circ} ) + sin ( 15 0 ^{\circ} ) - tan ( 3 0 ^{\circ} ) cos ( 15 0 ^{\circ} )}{cos ( 15 0 ^{\circ} )}

Use the following trivial identity:

sin (15 0^{\circ}) = \frac{1}{2}

Use the following trivial identity:

cos (15 0^{\circ}) = - \frac{3}{2}

Use the following trivial identity:

tan (3 0^{\circ}) = \frac{3}{3}

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

Simplify

\frac{\frac{1}{2} ( - \frac{3}{2} ) + ( - \frac{3}{2} ) ^{2} + \frac{1}{2} - \frac{3}{3} ( - \frac{3}{2} )}{- \frac{3}{2}} : - \frac{7 3 - 3}{6}

= - \frac{7 3 - 3}{6}

tan (\frac{\frac{3}{2}}{- \frac{1}{2}})

sin^{4} (\frac{π}{3})

sin^{2} (3 0^{\circ}) - cot^{2} (6 0^{\circ})

1 + 2 cos (6 0^{\circ})

sin (\frac{6 π}{12})

What is the value of sin(150)+cos(510)+tan(4110)-tan(210) ?
The value of sin(150)+cos(510)+tan(4110)-tan(210) is -(7sqrt(3)-3)/6