如何證arctan(s)+arctan(1/s)=pai/2

For 0 (less than) arctan(s), arctan(1/s) (less than) pi/2,

tan [pi/2 - arctan(1/s)] = 1 / tan [arctan(1/s)] = 1/(1/s) = s
pi/2 - arctan(1/s) = arctan(s)
arctan(s) + arctan(1/s) = pi/2

In general,
tan x = s
x = arctan(s) = y + npi
where tan y = s and n is an integer.
We will not have arctan(s) + arctan(1/s) = pi/2.

For -pi/2 (less than) arctan(s), arctan(1/s) (less than) 0,
replace s by -s in the above yield
arctan(s) + arctan(1/s) = -pi/2

謝謝您提供的意見^_^

2010-01-13 12:00:38 補充：
我認為在For -pi/2 (less than) arctan(s), arctan(1/s) (less than) 0的條件後
直接註明：當s<0時(我不懂replace s by -s in the above yield怎麼代換),
再進行推導可不可以？
變成：tan [-pi/2 - arctan(s)] = 1 / tan [arctan(s)] = 1/s
-pi/2 - arctan(s) = arctan(1/s)
所以arctan(s) + arctan(1/s) = -pi/2

題目錯了!
s>0時 arctan(s)+arctan(1/s)= pi/2
s<0時,arctan(s)+arctan(1/s)= -pi/2