What is the value of tan(-225)+tan(405) ?

The value of tan(-225)+tan(405) is 0

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tan(-225)+tan(405)

tan (- 22 5^{\circ}) + tan (40 5^{\circ})

0

Solution steps

tan (- 22 5^{\circ}) + tan (40 5^{\circ})

Use the following property:

tan (- x) = - tan (x)

tan (- 22 5^{\circ}) = - tan (22 5^{\circ})

= - tan (22 5^{\circ}) + tan (40 5^{\circ})

tan (22 5^{\circ}) = tan (4 5^{\circ})

tan (22 5^{\circ})

Rewrite

22 5^{\circ}

18 0^{\circ} + 4 5^{\circ}

= tan (18 0^{\circ} + 4 5^{\circ})

Apply the periodicity of

tan

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

tan (18 0^{\circ} + 4 5^{\circ}) = tan (4 5^{\circ})

= tan (4 5^{\circ})

= - tan (4 5^{\circ}) + tan (40 5^{\circ})

tan (40 5^{\circ}) = tan (4 5^{\circ})

tan (40 5^{\circ})

Rewrite

40 5^{\circ}

18 0^{\circ} \cdot 2 + 4 5^{\circ}

= tan (18 0^{\circ} 2 + 4 5^{\circ})

Apply the periodicity of

tan

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

tan (18 0^{\circ} \cdot 2 + 4 5^{\circ}) = tan (4 5^{\circ})

= tan (4 5^{\circ})

= - tan (4 5^{\circ}) + tan (4 5^{\circ})

Simplify

= 0

\frac{sin ( 4 0 ^{\circ} )}{sin ( 2 0 ^{\circ} )}

\frac{- 313.3}{sin ( 6 0 ^{\circ} )}

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

\frac{1 + sin ( \frac{π}{2} )}{1 - cos ( 2 π )}

8 sin (1 5^{\circ})