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8.
(a)
tan105° = tan(45°+60°)
= (tan45°+tan60°)/[1-tan45°tan60°]
tan45° = 1
tan60° = √3
tan105° = (1+√3)/[1-(1)(√3)]
= (1+√3)/(1-√3)
= (1+√3)/(1-√3) x (1+√3)/(1+√3)
= (1+2√3+3)/(1-3)
= (4+2√3)/(-2)
= -(2+√3)
(b)
sin²75° = ½(1-cos150°)
= ½(1+cos30°) [ note: cos30° = ½√3 ]
= ½(1+½√3)
= ¼(2+√3)
(a)
tan105° = tan(45°+60°)
= (tan45°+tan60°)/[1-tan45°tan60°]
tan45° = 1
tan60° = √3
tan105° = (1+√3)/[1-(1)(√3)]
= (1+√3)/(1-√3)
= (1+√3)/(1-√3) x (1+√3)/(1+√3)
= (1+2√3+3)/(1-3)
= (4+2√3)/(-2)
= -(2+√3)
(b)
sin²75° = ½(1-cos150°)
= ½(1+cos30°) [ note: cos30° = ½√3 ]
= ½(1+½√3)
= ¼(2+√3)
Date Posted:
5 years ago