Reciprocal relations of trigonometric ratios are explained here to represent the relationship between the three pairs of trigonometric ratios as well as their reciprocals.
Let OMP be a right angled triangle at M and ∠MOP = θ.
According to the definition of trigonometric ratios we have,
● sin θ = perpendicular/hypotenuse = MP/PO ………….. (i)
and csc θ = hypotenuse/perpendicular = PO/MP ………….. (ii)
From (i) sin θ = 1/(PO/MP)
⇒ sin θ = 1/csc θ ………………… (A)
Again, from (ii) csc θ = 1/(MP/PO)
⇒ csc θ = 1/sin θ ………………… (B)
From (A) and (B) we conclude that
sin θ and csc θ are reciprocal of each other.
● cos θ = adjacent/hypotenuse = OM/OP ………….. (iii)
and sec θ = hypotenuse/adjacent = OP/OM ………….. (iv)
From (iii) cos θ = 1/(OP/OM)
⇒ cos θ = 1/sec θ ………………… (C)
Again, from (iv) sec θ = 1/(OM/OP)
⇒ sec θ = 1/cos θ ………………… (D)
From (C) and (D) we conclude that
cos θ and sec θ are reciprocal of each other.
● tan θ = perpendicular/adjacent = MP/OM ………….. (v)
and cot θ = adjacent/perpendicular = OM/MP ………….. (vi)
From (v) tan θ = 1/(OM/MP)
⇒ tan θ = 1/cot θ ………………… (E)
Again, from (vi) cot θ = 1/(MP/OM)
⇒ cot θ = 1/tan θ ………………… (F)
From (E) and (F) we conclude that
tan θ and cot θ are reciprocal of each other.
To find values of trig functions we can use these reciprocal relationships to solve different types of problems.
Note:
From the above discussion about the reciprocal trigonometric functions we get;
1. sin θ ∙ csc θ = 1
2. cos θ ∙ sec θ = 1
3. tan θ ∙ cot θ = 1
● Trigonometric Functions
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