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Mathematics 15 Online
OpenStudy (kaylala):

Find the value of the following using the concepts of both addition and subtraction formulas and reference angle: tan (-255)

OpenStudy (kaylala):

@eliassaab

OpenStudy (***[isuru]***):

tan ( -225 ) = - tan ( 225) =- tan ( phi + 75 ) = - tan 75 =- tan ( 45 + 30)

OpenStudy (***[isuru]***):

\[- \tan ( 30 + 45) =- \frac{ \sin(30 + 45) }{ \cos (30 + 45) }\]\[- \tan (30 + 45) = - \frac{ \sin30 \ \cos45 + \cos30 \ \sin45 }{ \cos 30 \ \cos 45 - \sin30 \ \sin 45 }\]

OpenStudy (***[isuru]***):

\[- \tan ( 45 + 30) = -\frac{ (\frac{ 1 }{ 2} \times \frac{ 1 }{ \sqrt{2} }) + (\frac{ \sqrt{3} }{ 2 }\times \frac{ 1 }{ \sqrt{2} })}{(\frac{ \sqrt{3} }{ 2 }\times \frac{ 1 }{ \sqrt{2} })-(\frac{ 1 }{ 2 }\times \frac{ 1 }{ \sqrt{2} } ) }\]

OpenStudy (***[isuru]***):

\[- \tan 75 = \frac{ 1 + \sqrt{3} }{ \sqrt{3} - 1} =- \frac{ ( 1 + \sqrt{3})(\sqrt{3} +1) }{ (\sqrt{3} -1)(\sqrt{3} +1) } =- \frac{ 1 +3 + 2\sqrt{3} }{ 2} = - 2 + \sqrt{3} = \tan(-225)\]

OpenStudy (***[isuru]***):

\[\tan (-225) = -( 2 + \sqrt{3} )\]

OpenStudy (***[isuru]***):

got it ?

OpenStudy (kaylala):

yes got it. thanks @***[ISURU]***

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