唔該..有邊個可以計個步驟同個ans比我睇？
prove the following identities
(Asec x+Btan x)^2-(Atan x+Bsec x)^2=A^2-B^2
數學高手..幫幫我..
2008-11-14 5:15 am
回答 (2)
2008-11-14 5:38 am
✔ 最佳答案
Explanation as follows:圖片參考：http://i238.photobucket.com/albums/ff245/chocolate328154/Maths547.jpg?t=1226583479
2008-11-13 23:35:15 補充：
Note: 1 + tan^2 x = sec^2 x
That comes the last few steps.
參考: My Maths Knowledge
2008-11-14 6:19 am
(Asec x + Btan x)^2 - (Atan x + Bsec x)^2
= [(Asec x + Btan x) + (Atanx + Bsec x)][(Asec x + Btan x) - (Atanx + Bsec x)]
= [A(sec x + tan x) + B(tanx + sec x)][(A(sec x - tan x) - B(sec x - tan x)]
= A^2(sec x + tan x)(sec x - tan x) - B^2(tan x + sec x)(sec x - tan x)
= A^2[ sec^2(x) - tan^2(x)] - B^2[sec^2(x) - tan^2(x)]
= A^2 - B^2 since sec^2(x) - tan^2(x) = 1.
= [(Asec x + Btan x) + (Atanx + Bsec x)][(Asec x + Btan x) - (Atanx + Bsec x)]
= [A(sec x + tan x) + B(tanx + sec x)][(A(sec x - tan x) - B(sec x - tan x)]
= A^2(sec x + tan x)(sec x - tan x) - B^2(tan x + sec x)(sec x - tan x)
= A^2[ sec^2(x) - tan^2(x)] - B^2[sec^2(x) - tan^2(x)]
= A^2 - B^2 since sec^2(x) - tan^2(x) = 1.
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