How to identify sides?Identify the angle with respect to which the tratios have to be calculated. Sides are always labelled with respect to the ‘θ’ being considered.
Let us take two cases:
In a right triangle ABC, rightangled at B. Once we have identified the sides, we can define six tRatios with respect to the sides.
Note from above six relationships:
However, it is very tedious to write full forms of tratios, therefore the abbreviated notations are:
Example: If in a rightangled triangle ABC, rightangled at B, hypotenuse AC = 5cm, base BC = 3cm and perpendicular AB = 4cm and if ∠ACB = θ, then find tan θ, sin θ and cos θ.
Solution:
Given,
In ∆ABC,
Hypotenuse, AC = 5cm
Base, BC = 3cm
Perpendicular, AB = 4cm
Then,
tan θ = Perpendicular/Base = 4/3
Sin θ = Perpendicular/Hypotenuse = AB/AC = ⅘
Cos θ = Base/Hypotenuse = BC/AC = ⅗
An equation involving trigonometric ratio of angle(s) is called a trigonometric identity, if it is true for all values of the angles involved. These are:
Example: Express the ratios $\backslash cos\; A$cosA, $\backslash tan\; A$tanA, and $\backslash sec\; A$secA in terms of $\backslash sin\; A$sinA.
Solution: Since $\backslash cos^2\; A\; +\; \backslash sin^2\; A\; =\; 1$cos^{2}A + sin^{2}A = 1, therefore:
cos^{2}A = 1 − sin^{2}A
i.e., cosA = ±√1−sin^{2}A
This gives:
cosA = √1 − sin^{2}A
Hence,
and
Note: A trigonometric ratio only depends upon the angle ‘θ’ and stays the same for same angle of different sized right triangles.
Values of Trigonometric Ratios of Specified Angles
The value of sin θ and cos θ can never exceed 1 (one) as opposite side is 1. Adjacent side can never be greater than hypotenuse since hypotenuse is the longest side in a rightangled ∆.
Example: If tan θ + cot θ = 5, find the value of tan2θ + cotθ.
Solution:
tan θ + cot θ = 5 … [Given
tan^{2}θ + cot^{2}θ + 2 tan θ cot θ = 25 … [Squaring both sides
tan^{2}θ + cot^{2}θ + 2 = 25
∴ tan^{2}θ + cot^{2}θ = 23
Example: If sec 2A = cosec (A – 27°) where 2A is an acute angle, find the measure of ∠A.
Solution:
sec 2A = cosec (A – 27°)
cosec(90° – 2A) = cosec(A – 27°) …[∵ sec θ = cosec (90° – θ)
90° – 2A = A – 27°
90° + 27° = 2A + A
⇒ 3A = 117°
∴ ∠A = 117°/3 = 39°
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