Plane trigonometry, by S.L. Loney.

TRIGONOMETRY.

39. To find the relations between the trigonometricat
ratios of two complementary angles.

Let the revolving line, starting from
any acute angle AOP, equal to
0. From any point P on it
draw PM perpendicular to OA.
Since the three angles of a
triangle are together equal to
two right angles, and since OMP
is a right angle, the sum of the o
two angles MOP and OPM is a
right angle.
They are therefore complementary and

OA, trace out,

900
M A.

z OPM = 90~ -.
[When the angle OPM is considered, the line PM is
the " base" and MO is the "perpendicular."]
We then have
MO
sin (90~- ) = sin MPO= =cos A OP = cos,
PM
cos (90~ - ) = cos MPO = = sin A OP = sin 0,
MO
tan (90 - 0) = tan MPO = = cot AOP = cot 0,
PM
cot (90~ - 0) = cot MPO = - = tan A OP = tan 0,
MO
PO
cosec (90~ - 0) = cosec MPO =M = sec A OP = sec 0,
PO
and sec (900 - 0) = sec MPO = = cosec A OP = cosec O.
PM