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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