Sec. 10.10 Angles between lines 173
10.10.6 Addition formulae for tangents of duo-angles .....
(i) We first note that ifαd,βd∈DA(F)andαd+βd= (^90) dF, neither duo-angle
being null or right, then tanαdtanβd=1. For we have thatx 6 =0, so that by (10.10.2)
x 4 x 5 −y 4 y 5 =0 and thus
y 4
x 4
y 5
x 5
= 1.
(ii) Next we note that, as tanαd=sinαd/cosαd, it follows from the above addi-
tion formulae for cosine and sine that
tan(αd+βd)=tanαd+tanβd
1 −tanαdtanβd,
provided that 0 does not occur in a denominator, that is provided none ofαd,βd,αd+
βdis a right duo-angle; this can be done separately for the cases considered in 10.10.5.
In fact this addition formula for the tangent function can be verified without subdivi-
sion into cases, as
y 6
x 6
=− 2
(y 5 −y 4 )(x 5 −x 4 )
(y 5 −y 4 )^2 −(x 5 −x 4 )^2,
and we wish to show that this is equal to
y 5 /x 5 +y 4 /x 4
1 −y 4 y 5 /x 4 x 5=
x 4 y 5 +x 5 y 4
x 4 x 5 −y 4 y 5. (10.10.3)
On subtracting the first of these expressions from the second, we obtain a quotient
the numerator of which is equal to
(x 4 y 5 +x 5 y 4 )[(y 5 −y 4 )^2 −(x 5 −x 4 )^2 ]+ 2 (x 4 x 5 −y 4 y 5 )(y 5 −y 4 )(x 5 −x 4 )
=(x 4 y 5 +x 5 y 4 )[y^25 +y^24 −x^25 −x^24 + 2 (x 4 x 5 −y 4 y 5 )]
+ 2 (x 4 x 5 −y 4 y 5 )[x 4 y 4 +x 5 y 5 −(x 4 y 5 +x 5 y 4 )]
=(x 4 y 5 +x 5 y 4 )[y^25 +y^24 −x^25 −x^24 ]+ 2 (x 4 x 5 −y 4 y 5 )(x 4 y 4 +x 5 y 5 )
=(x 4 y 5 +x 5 y 4 )[y^25 +y^24 −x^25 −x^24 ]+ 2 (x^24 x 5 y 4 −y^24 x 4 y 5 +x^25 x 4 y 5 −y^25 x 5 y 4 )
=(x^25 +y^25 −x^24 −y^24 )(x 4 y 5 −x 5 y 4 )= 0 ,asx^24 +y^24 =x^25 +y^25 =k^2. This identity then implies the standard addition formula for
the tangents of duo-angles.
(iii) We also wish to show that
tan(αd+ (^90) dF)=
− 1
tanαd,
whenαdis neither null nor right. For withx 5 = 0 ,y 5 =k(10.10.1) gives
x 6 =k(k−y 4 )^2 −x^24
x^24 +(k−y 4 )^2,y 6 = 2 k−x 4 (k−y 4 )
x^24 +(k−y 4 )^2