Plane trigonometry, by S.L. Loney.

TANGENT OF THE SUM OF TWO ANGLES. 101
7?P OQ
But, since the angles RPN and NOQ are equal, we have P= OQ
tN ON
PR PN
and therefore - PN= tan B.
OQ ON
tan A - tan B tan A - tan B
Hence tan (A -/B) = + tan RPN tan B 1 + tan A tan B
100. As particular cases of the preceding formulae,
we have, by putting B equal to 45~,
tan A + 1 1 + tan A
tan (A + 450) = tan A + _ 1 tan A
1 -tan A 1 -tan A'
andi~~ i /A Ao tan A - 1
and tan (A - 45o) = tln A A.
1 + tan A'
Similarly as in Art. 98 we may prove that
cot A cot B - 1
cot (A + B) —
cot A + cot B
cot A cot B + I
and cot (A - B) cot B-cot A
tan 450 + tan 300
101. Ex. 1. tan 750 = tan (45~ + 300) = ltan450tan 30
1
3_ /3 +1 (,3 + 1)2 4 + 2/3
1 __1_,3-1 -3-1 - 2 2+3
J/3
=2+ 1-73205.. = 3-73205....
tan 450 - tan 30~
Ex. 2. tan 15~= tan (45o - 30~) = tan 45 tan 30
1
1 + tan 45~ tan 30~,/3 =3 -1 (_3- 1) 4 - 2v323
1 -,33+-1 3-1 - 2 =2-J3
1+1,/3
=2-1-73205...=-26795....