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whence
Z
yy
Z
xy
=
t
+
s
tan
t
+
tan
s
tan
x
=
st
=
t
=
tan(
t
+
s
)
,
1
−
1
−
tan
s
tan
(3
.
32)
Z
xx
Z
yx
=
t
−
s
tan
t
−
tan
s
tan
y
=−
st
=
t
=
tan(
t
−
s
)
,
1
+
1
+
tan
s
tan
and hence
x
=
t
+
s
y
=
t
−
s
,
(3
.
33)
=
x
+
y
2
x
−
y
2
t
s
=
.
Thus, we have simple arithmetic relations between
t
,
s
.
Furthermore, it is easy to show that the Groom-Bailey decomposition
x
,
y
and
1
1
s
1
−
+
t
a
0
1
1
1
[
e
]
=
g
√
1
√
1
√
1
t
2
s
2
a
2
+
+
+
t
1
s
1
01
−
a
and the Bahr decomposition
cos
e
x
x
−
sin
y
0
[
e
]
=
sin
x
cos
y
0
e
y
are identical (Smith, 1995). Really, in view of (3.33),
cos
e
x
x
−
sin
y
0
=
[
e
]
sin
x
cos
y
0
e
y
cos
cos
e
x
t
−
sin
t
s
sin
s
0
=
sin
t
cos
t
sin
s
cos
s
0
e
y
1
1t n
1
(3
.
34)
−
tan
t
s
+
a
0
=
e
cos
t
cos
s
tan
t
1
tan
s
1
01
−
a
1
1
s
1
−
+
t
a
0
1
1
1
=
g
√
1
√
1
√
1
,
t
2
s
2
a
2
+
+
+
t
1
s
1
0
1
−
a
where
e
x
+
e
y
e
x
+
e
y
e
x
−
e
y
e
=
,
g
=
,
a
=
e
y
.
2
2
e
x
+
