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secondary 4 | A Maths
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Charlotte Lim
Charlotte Lim

secondary 4 chevron_right A Maths

Pls help with both parts. Thanks!!

Date Posted: 2 months ago
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snell
Snell's answer
179 answers (A Helpful Person)
1st
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)