a.maths,,help

2007-11-13 4:09 am
(1-tanX)^2+(1-COTX)^2-(secX-COsecX)^2=0

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回答 (2)

2007-11-13 4:28 am
✔ 最佳答案
L.H.S. = (1-tanx)^2+(1-cotx)^2-(secx-cosecx)^2

=1-2tanx+tan^2 x+1-2cotx+cot^2x-sec^2x+2secxcosecx-cosec^2x

=sec^2x+cosec^2x-2tanx-2cotx-sec^2x+2secxcosecx-cosec^2x

=-2tanx-2cotx+2secxcosecx

=-2(sinx/cosx+cosx/sinx)+2/sinxcosx

=-2[(sin^2x+cos^2x)/sinxcosx]+2/sinxcosx

=-2/sinxcosx+2/sinxcosx

=0

=R.H.S.


2007-11-13 22:06:10 補充:
Reminder: 1 + tan^2 x = sec^2 x 1 + cot^2 x = cosec^2 x
參考: My Maths Knowledge
2007-11-13 7:33 pm
(1-tanx)^2 + (1-cotx)^2 - (secx-cosecx)^2 = 0
=> (1-2tanx+tan^2 x) + (1-2cotx+cot^2 x) - (sec^2 x - 2cosecx/secx + cosec^2 x) = 0
=> (sec^2 x - 2tanx) + (cosec^2 x - 2cotx) - (sec^2 x - 2cotx + cosec^2 x) = 0
=> -2tanx = 0
=> tan x = 0
=> x = 180n or (n*pi in radian) where n = 0, 1, 2, ...


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