201] 
SECOND NOTE ON THE THEORY OF LOGARITHMS. 
223 
Hence, attending to the equation 
tan -1 /3 — tan -1 a — tan -1 a _ + e" nr, 
l+a/3 
where, when l+a/3 is positive, e" is equal to zero; but when l+a/3 is negative, 
e" is equal to +1 or — 1, according as /3 — a is positive or negative: or what is the 
same thing (a, /3 being of opposite signs when l+a/3 is negative), 
we find 
1 + a/3 = + , e"' = 0, 
1 + a/3 = —, e" = + 1 = /3 — a = /3 = — a, 
E = e — e + e" — e", 
where e, e, e", e" are defined by the conditions 
X = +, 
e =0, 
X = — , 
Hi 
II 
1+ 
h-‘ 
III 
x' = + , 
6' =0, 
x' =-, 
e' =±1=2/', 
xx' + yy' = + , 
e" =0, 
xx' + yy' = - , 
e 7/ =± 1 =xy' 
1+^ = +, e"' = 0, 
1 + 
X X 
y t 
X x' 
a- \ -.y=y- = -.y 
— 1 — / _ — / — _ 
xxx 
Suppose, to fix the ideas, that x, y are each of them positive, we have e = 0; 
and considering the several cases : 
1. x' = + , y' = + . 
Here xx + yy' = +, 1 + - ^ = + ; and consequently not only e' = 0, but also e" = 0, 
= 0, and thence E = 0. 
2. x = +, y' = ~. 
Here e = 0. Moreover, xx' being positive, xx' + yy' and 1 + ^ ~, will have the 
same sign. If they are both positive, then e" = 0, e" = 0; but if they are both 
negative, then 
e" = + 1 = xy' — x'y = - 
(since xx' = +) and e"' = ± 1 = ^, i.e. e" = e"'. Hence in either case we have E= 0.