What is the value of sin(tan(50))+sin(tan(310)) ?

The value of sin(tan(50))+sin(tan(310)) is 0

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sin(tan(50))+sin(tan(310))

sin (tan (5 0^{\circ})) + sin (tan (31 0^{\circ}))

0

Solution steps

sin (tan (5 0^{\circ})) + sin (tan (31 0^{\circ}))

tan (31 0^{\circ}) = tan (13 0^{\circ})

tan (31 0^{\circ})

Rewrite

31 0^{\circ}

18 0^{\circ} + 13 0^{\circ}

= tan (18 0^{\circ} + 13 0^{\circ})

Apply the periodicity of

tan

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

tan (18 0^{\circ} + 13 0^{\circ}) = tan (13 0^{\circ})

= tan (13 0^{\circ})

= sin (tan (5 0^{\circ})) + sin (tan (13 0^{\circ}))

Rewrite using trig identities:

tan (5 0^{\circ}) = - tan (13 0^{\circ})

tan (5 0^{\circ})

Use the following identity:

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

tan (x)

Use the following property:

tan (θ) = - tan (- θ)

tan (x) = - tan (- x)

= - tan (- x)

Apply the periodicity of

tan

tan (18 0^{\circ} + θ) = tan (θ)

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

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

= - tan (36 0^{\circ} - 5 0^{\circ})

Simplify

= - tan (31 0^{\circ})

tan (31 0^{\circ}) = tan (13 0^{\circ})

tan (31 0^{\circ})

Rewrite

31 0^{\circ}

18 0^{\circ} + 13 0^{\circ}

= tan (18 0^{\circ} + 13 0^{\circ})

Apply the periodicity of

tan

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

tan (18 0^{\circ} + 13 0^{\circ}) = tan (13 0^{\circ})

= tan (13 0^{\circ})

= - tan (13 0^{\circ})

= sin (- tan (13 0^{\circ})) + sin (tan (13 0^{\circ}))

Use the following property:

sin (- x) = - sin (x)

sin (- tan (13 0^{\circ})) = - sin (tan (13 0^{\circ}))

= - sin (tan (13 0^{\circ})) + sin (tan (13 0^{\circ}))

Simplify

= 0

sec (6 4^{\circ})

cos (6) 2

cos (\frac{- 2}{2})

\frac{8}{sin ( 16. 5 ^{\circ} )}

arcsec (\frac{41}{9})

What is the value of sin(tan(50))+sin(tan(310)) ?
The value of sin(tan(50))+sin(tan(310)) is 0