A-maths~

1. sin^2 A/(1+2cos^2 A)=3/19
Try to reduce it into an equation with SINGLE FUNCTION.
(sin A)^2/[1+2(cos A)^2]=3/19
(sin A)^2/[1+2(1 - (sin A)^2)]=3/19
(sin A)^2/[3 - 2(sin A)^2)]=3/19
19 (sin A)^2 = 3 * [3 - 2(sin A)^2)]
19 (sin A)^2 = 9 - 6(sin A)^2
25 (sin A)^2 = 9
(sin A)^2 = 9/25
sin A = -3/5 since A is in the 3rd quadrant
=> sec A = 1/cos A = -5/4, tan A = 3/4
tan A/(1 + sec A) = 3/4 / (1 - 5/4)
= 3/4 / (-1/4)
= -3

2. sin(750度+a)tan(765度+a)tan(225度-a)sec(a-60度)=1
= sin(2*180 + 30 + a) * [tan(2*180 + 45 + a) * tan(225 - a)] / cos (60 - a)
= sin(30 + a) / cos (60 - a) * [tan(45 + a) * tan(225 - a)]
= [(sin 30cos a + cos 30sin a) / (cos 60cos a + sin 60sin a)] * [(tan 45 + tan a)/(1 - tan 45tan a) * (tan 225 - tan a)/(1 + tan 225tan a)]
= [(1/2*cos a + 3^(1/2)/2*sin a) / (1/2*cos a + 3^(1/2)*sin a)] * [(1 + tan a)/(1 - tan a) * (1 - tan a)/(1 + tan a)]
= 1

做呢類數, 其實冇什麼特別的心得, 只要小心的做, 寫一步驟, 諗一諗有冇寫錯, 便是了.

重有做第一題做到sin A= -3/5, 其實可以自己找出 A 的數值(216.87度), 再快速代入sin^2A/1+2cos^2A, 檢查是否仍然等於3/19, 如果不同, 就顯然當中有錯. 最後得出-3, 也可以自己把216.87度代入tanA/(1+secA)去核對是否真係等於-3. 這樣就穩當得多啦.

當然formula要熟練啦, 要多練習思考才行.
可參考以下有關trigonometric identities的網址:
http://en.wikipedia.org/wiki/Trigonometric_identity