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(7) cot a
Hence, The co-tangent of any arc is equal to its co-sine divided by its sine.
(6) × (7) = (8) tan a cot a = 1.
sin (90° - a)
cos (90° -- a)
Hence, The tangent of any arc into its co-tangent is equal to 1.
(8) ÷ cot a = (9) tan a=
Hence, The tangent of any arc is equal to the reciprocal of its co-tangent.
(8) ÷ tan a = = (10)
sec (90° — a)
Hence, The co-tangent of any arc is equal to the reciprocal of its tangent.
cot a =
CM CO CT : CR, or cos a: 1 :: 1 : sec a.
... (11) sec a
(12) cosec a
Hence, The secant of any arc is equal to the reciprocal of its co-sine.
cos (90° - a)
Hence, The co-secant of any arc is equal to the reciprocal of its sine.
CR2 = CO2 + OR2,
(13) sec2 a = 1 + tan2 a.
Hence, The square of the secant of any arc is equal to 1, plus the square of its tangent.
... sec2 (90°— a) ·
4. vers a
Hence, The square of the co-secant is equal to 1, plus the square of the co-tangent.
(14) cosec2 a= 1 + cot2 a.
82. Summary of Fundamental Formulas.
1. sin2 a + cos2 a 1.
2. sin2 a
1- cos2 a.
3. cos2 a
1- sin2 a.
7. cot a
5. covers a =
6. tan a
-1- sin a.
8. tan a cot a
1 + tan2 (90°— a).
1. Prove that the above formulas become homogeneous by the introduction of R.
2. Deduce formulas (5), (7), (12) and (14) from the diagram.
3. Prove that the above formulas are true if a is in the second, third, or fourth quadrant.
85. Functions of Negative Arcs.
We first find the sine and co-sine of - a, in terms of the functions of a from the diagram. Then, dividing the sine by the co-sine, the cosine by the sine, taking the reciprocal of the cosine and the reciprocal of the sine, we have
86. Functions of (n 90° a).
2. Let n be 1 and a be positive.
tan (90° + a) sec (90° + a)
1. Let n be 1 and a be negative.
From the figure of the last article, and by similar
(90° — a) = sin a,
sin (90° — a) = cos a,
cos a, cot a,
3. Let n be 2, and a be negative. sin (180°-a) = tan (180°— a) sec (180°— a) =
(90° + a)
(90° + a) cosec (90° a) =
·tan a, cot (180° — a)
cos a, cot a,