Reciprocal Relations of Trigonometric Ratios

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.

Reciprocal Relations of Trigonometric RatiosReciprocal Relations of Trigonometric Ratios

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






10th Grade Math

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