272 A Demonstration of Le Gendre’s Theorem 
and, omitting the mer factors on both sides of each equation ; 
Cae a3) a 
#)(4 2 
0 B04 
a 
tan A 
3A 
=(1- a) (a3; tan A. 
and, by multiplying, and neglecting small quantities of the se- 
cond order ; 
2B a es tA be ' 
‘tanA  6r  tanA 6? 
(3) 
oC az. OA ct 
tan C 6 ~ ta A 6r2° 
Again, in the plane triangle, we have 
a’ = +c? — 2becosA 
= 
oo 
a* + c? — 2ac cos B, 
a* + b* — 2al cos C: 
but, if s represent the area of the triangle, then 2s= de sin A 
2 
=acsin B = al sin C: and hence Jc cos A = 
= —~_: gh cos C = 
tan B ” 
—a? ae cos B 
a The values of a?, b*, c* may there- 
fore be thus represented, viz. 
ps SOPs 
Qs 
tan A? 
fees a? + 6?-+-¢2 fay 284 
ity 2 tan B? 
ee a+h+¢ x Pome 
2 tan C 
Let these v es be substituted in the equations (3); then 
B s 3A s 
tan B = Sr? tan A Tteuts ae Sr? tan B? 
3C s 3A s 
in ! sttmA ana? StnC” 
and hence 
(2B - = 
aa) tan A= (2A — 
(2C— =) tanA = (0A — 
s 
=) tan B, 
Sre 
tion, viz. 
(4) 
$ 
=) tan C, 
Take the sum of these two equations, and of this identical equa- 
then, 
(2A — sa) tanA = (A-<) tan A 
(#A+8B48C——_) tan A=(2A— =; ) (tan A + tan B # 
tan C). 
Nows 
