RECIPROCAL RELATION OF TRIGONOMETRIC RATIOS

In the six trigonometric ratios sin, cos, tan, csc, sec and cot, there is a reciprocal relation among them.

Here, the pairs of trigonometric relations are given between which we have reciprocal relation.

sinθ <----> cscθ

cosθ <----> secθ

tanθ <----> cotθ

That is,

 sinθ = 1/cscθcosθ = 1/secθtanθ = 1/cotθtanθ = sinθ/cosθ cscθ = 1/sinθsecθ = 1/cosθcotθ = 1/tanθcotθ = cosθ/sinθ

To have better understanding on reciprocal relations of trigonometric ratios, first we have to know the shortcut SOHCAHTOA which is related to the trigonometric ratios sin, cos and tan.

To understand the shortcut, first we have to divide SOHCAHTOA in to three parts as given below. What do SOH, CAH and TOA stand for ? From the above figures, we can derive formulas for the three trigonometric ratios sin, cos and tan as given below. Solved Problems

Problem 1 :

In the right triangle PQR shown below, find the values of sinθ, cosθ and tanθ. Using those values, find the values of cscθ, secθ and cotθ. Solution :

From the triangle shown above,

opposite side = 5

hypotenuse = 13

Therefore,

 sinθ = PQ/RQ = 5/13cosθ = PR/RQ = 12/13tanθ = PQ/PR = 5/12 cscθ = 1/sinθ = 13/5secθ = 1/cosθ = 13/12cotθ = 1/tanθ = 12/5

Problem 2 :

In the right triangle ABC shown below, find the values of sinθ, cosθ and tanθ. Using those values, find the values of cscθ, secθ and cotθ. Solution :

From the right triangle shown above,

AC = 24

BC = 7

By Pythagorean Theorem,

AB2 = BC2 + CA2

AB2 = 72 + 242

AB2 = 49 + 576

AB2 = 625

AB2 = 252

AB = 25

Now, we can use the three sides to find the six trigonometric ratios of angle θ.

Therefore,

 sinθ = BC/AB = 7/25cosθ = AC/AB = 24/25tanθ = BC/AC = 7/24 cscθ = 1/sinθ = 25/7secθ = 1/cosθ = 24/25cotθ = 1/tanθ = 24/7

Problem 3 :

In triangle ABC, right angled at B, 15sinA = 12. Find the six trigonometric ratios of angle A and C.

Solution :

Given : 15sinA = 12.

sinA = 12/15

sinA = 4/5

Therefore,

opposite side = 4

hypotenuse = 5

Let us consider the right triangle ABC where right angled at B, with

BC = 12

AC = 15 By Pythagorean theorem,

AC2 = AB2 + BC2

52 = AB2 + 42

25 = AB2 + 16

25 - 16 = AB2

9 = AB2

3= AB2

3 = AB

Now, we can use the three sides to find the five trigonometric ratios of angle A and six trigonometric ratios of angle C.

Therefore,

 sinA = 4/5cosA = AB/AC = 3/5tanA = BC/AB = 4/3sinC = AB/AC = 3/5cosC = BC/AC = 4/5tanC = AB/BC = 3/4 cscA = 1/sinA = 5/4secA = 1/cosA = 5/3cotA = 1/tanA = 3/4cscC = 1/sinC = 5/3secC = 1/cosC = 5/4cotC = 1/tanC = 4/3

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