**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.

sin θ <---------> csc θ

cos θ <---------> sec θ

tan θ <---------> cot θ

More clearly,

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 ?

Here is the answer

From the above figures, we can derive formulas for the three trigonometric ratios sin, cos and tan as given below.

csc θ = Hypotenuse / Opposite side

sec θ = Hypotenuse / Adjacent side

cot θ = Adjacent side/Opposite side

**Problem 1 :**

In the right triangle PQR given below, find the six trigonometric ratios of the angle θ

**Solution :**

From the figure given above,

opposite side = 5

adjacent side = 12

hypotenuse = 13

Once we get get the values of sinθ, cosθ and tanθ, we will be able to get the values of other trigonometric ratios cscθ, secθ and cotθ using reciprocal relation.

Therefore,

**Problem 2 :**

From the figure given below, find the six trigonometric ratios of the angle θ.

**Solution : **

From the figure given above, AC = 24 and 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 and find the trigonometric ratios sin, cos and tan of angle θ.

Once we get get the values of sinθ, cosθ and tanθ, we will be able to get the values of other trigonometric ratios cscθ, secθ and cotθ using reciprocal relation.

Therefore,

**Problem 3 :**

In triangle ABC, right angled at B, 15 sin A = 12. Find the other five trigonometric ratios of the angle A. Also find the six ratios of the angle C

**Solution : **

Given that 15 sin A = 12, so sin A = 12 / 15

Therefore, opposite side = 12 and hypotenuse = 15

Let us consider the triangle ABC where right angled at B, with BC = 12 and AC = 15.

By Pythagorean theorem,

AC² = AB² + BC²

15² = AB² + 12²

AB² = 15² - 12²

AB² = 225 - 144

AB² = 81

AB² = 9²

AB = 9

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

Therefore,

After having gone through the stuff given above, we hope that the students would have understood "Reciprocal relation of trigonometric ratios"

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