Plane trigonometry, by S.L. Loney.

122 TRIGONOMETRY.
Now tan 7~~ is positive so that we must take the upper sign.
Hence tan 71 = + (/6 - /2) - 1
2 -- /3
= (/6 - 2 - 1) (2 +,/3) ==/6 - /3 + /,2 - 2 = (/3 -,2) (/2 - 1).
15~
Since tan 15~= tan 195~, the equation which gives us tan - in terms
195~
of tan 15~ may be expected to give us tan - in terms of tan 195~. In
fact the value obtained from (1) by taking the negative sign before the
1950
radical is tan -2
Hene an1950 _ -8- 43 - 1 - (d - 2) - 1
Hence tan 2- = = 2 — /3
2 2-d/3 2-NaJ3
= ( - 6 +2 - 1) (2 + V3)= - (3 + 2) (/2 + 1),
so that -cot 7~~ = tan 971~ = - (/3 + 2) (^2 + 1).
A
#119. To explain why there is ambiguity when tan A
is found from the value of tan A.
We know, by Art. 84, that, if n be any integer,
tan (n7r + A) = tan A = k (say).
Hence any equation which gives us tan - in terms of k
ncr + A
may be expected to give us tan + also.
First, let n be even and equal to 2m.
Then
n7 r+A 2m7r+A / A\
tan -- - = tan 2 = tan t z7r + A
2 2 2I
A
= tan, as in Art. 84.
econdly, let n be odd and equal to p
Secondly, let n be odd and equal to 2p + 1.