We know that six trigonometric ratios can be formed using the three lengths a, b and c of sides of a right triangle ABC.
Interestingly, these ratios lead to the definitions of six basic trigonometric functions.
First, let us recall the trigonometric ratios which are defined with reference to a right triangle.
sin θ = opposite side/hypotenuse
cos θ = adjacent side/hypotenuse
With the help of sin θ and cos θ, the remaining trigonometric ratios tan θ, cot θ, csc θ and sec θ are determined by using the relations
tan θ = sin θ/cos θ
csc θ = 1/sin θ
sec θ = 1/cos θ
cot θ = cos θ/sin θ
And also,
sin θ = 1/csc θ
cos θ = 1/sec θ
Note :
1. sin θ and csc θ are reciprocal to each other.
2. cos θ and sec θ are reciprocal to each other.
3. tan θ and cot θ are reciprocal to each other.
Example :
1. If sin θ = 3/5, then csc θ = 5/3.
2. If cos θ = 4/5, then sec θ = 5/4.
3. If tan θ = 3/4, then cot θ = 4/3.
Problem 1 :
In the right triangle PQR given below, find the basic trigonometric ratios of the angle θ.
Solution :
In the triangle shown above, for the angle θ,
opposite side = 5
adjacent side = 12
hypotenuse = 13
Then, the basic trigonometric ratios of the angle θ are
sin θ = 5/13 cos θ = 12/13 tan θ = 5/12 |
csc θ = 13/5 sec θ = 13/12 cot θ = 5/12 |
Problem 2 :
From the figure given below, find the six trigonometric ratios of the angle θ.
Solution :
In the triangle shown above, by Pythagorean Theorem,
AB2 = BC2 + CA2
AB2 = 72 + 242
AB2 = 49 + 576
AB2 = 625
AB2 = 252
AB = 25
In the triangle shown above, for the angle θ,
opposite side = 7
adjacent side = 24
hypotenuse = 25
Then, the basic trigonometric ratios of the angle θ are
sin θ = 7/25 cos θ = 24/25 tan θ = 7/24 |
csc θ = 25/7 sec θ = 25/24 cot θ = 24/7 |
Problem 3 :
If sin θ = 13/85 and cos θ = 84/85, then find the values of tan θ and cos θ.
Solution :
Finding the value of tan θ :
tan θ = sin θ/cos θ
tan θ = (13/85)/(84/85)
tan θ = (13/85) ⋅ (85/84)
tan θ = (13 ⋅ 85)/(85 ⋅ 84)
tan θ = 13/84
Finding the value of cot θ :
cot θ = 84/13
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