Trigonometry is a study of the properties of triangles and trigonometric functions. It is used a lot in engineering, science, for building video games, and more. It deals with the relationship between ratios of the sides of a right-angled triangle with its angles. These ratios which are used for the study of this relationship are called Trigonometric ratios.
right-angled triangle
Trigonometric Ratios
There are six basic trigonometric ratios that establish the correlation between sides of a right triangle with the angle. If θ is the angle formed between the base and hypotenuse, in a right-angled triangle (as shown in the above figure), then
sin(θ) = Perpendicular/Hypotenuse
cos(θ) = Base/Hypotenuse
tan(θ) = Perpendicular/Base
The values of the other three function ,that is, cosec(θ), sec(θ), cot(θ) depend on sin(θ), cos(θ), tan(θ) respectively.
cot (θ) = 1/tan (θ) = Base/Perpendicular
sec (θ) = 1/cos (θ) = Hypotenuse/Base
cosec (θ) = 1/sin (θ) = Hypotenuse/Perpendicular
Trigonometric Identities
A trigonometric equation that holds for every viable value for an input variable on which it is defined is called trigonometric identities. One identity that we are familiar with is the Pythagorean Identity,
Sin2θ + Cos2θ = 1
Divide both sides of the Pythagorean Identity by cosine squared, which is allowed, then,
[cos2θ + sin2θ]/cos2θ = 1/cos2θ
cos2θ/cos2θ + sin2θ/cos2θ = 1/cos2θ
1 + sin2θ/cos2θ = 1/cos2θ
(using the definition tanθ = sinθ/cosθ)
1 + tan2θ = 1/cos2θ
(using the definition secθ = 1/cosθ)
1 + tan2θ = sec2θ
Therefore the next trigonometric identity is,
1 + tan2θ = sec2θ
Similarly, if divide both sides of the Pythagorean Identity by sine squared then we obtained the last identity,
cot2θ + 1 = cosec2θ
How to prove that sec2θ + csc2θ = sec2θ × csc2θ?
Proof:
To solve above problem we require below specified trigonometric identities and ratios :
sec(θ) = 1/cos(θ) and cosec(θ) = 1/sin(θ) ⇢ Eq. 1
sin2θ + cos2θ =1 ⇢ Eq. 2
sec2θ = 1 + tan2θ ⇢ Eq. 3
cosec2θ = 1+ cot2θ ⇢ Eq. 4
There are two ways to solve this problem
1. To prove LHS = RHS using identities
LHS = sec2(θ) + cosec2(θ)
=(1+ tan2θ) + (1+cosec2θ) (from 3 and 4)
=2 + tan2θ + cosec2θ
RHS= sec2θ × cosec2θ
=(1+tan2θ) × (1+cot2θ)
=2 + tan2θ + cot2θ
Therefore, LHS = RHS.
2. By using trigonometric ratios
LHS= sec2θ + cosec2θ
(from 1), [1/cos2θ ] + [1/sin2θ]
= [sin2θ + cos2θ] / [cos2θ × sin2θ]
(from 2), 1/[cos2θ × sin2θ]
= sec2θ × cosec2θ = RHS
Hence, the given trigonometric equation can be solved in two ways as mentioned above
Therefore, sec2(θ) + cosec2(θ) = sec2(θ) × cosec2(θ).
Similar Problems
Question 1: Prove, tan4(θ) + tan2(θ) = sec4(θ) – sec2 (θ) [Hint: take tan2(θ) as common]
Solution:
LHS= tan4θ + tan2θ = tan2θ (tan2θ + 1)
= (sec2θ – 1) (tan2θ + 1) {since, tan2θ = sec2θ – 1}
=(sec2θ – 1) sec2θ {since, tan2θ + 1 = sec2θ}
=sec4(θ) – sec2(θ) = RHS
Hence proved.
Question 2: Prove, cos θ / [(1 – tan θ)] + sin θ / [(1 – cot θ)] = sin θ + cos θ
Solution:
LHS = cos θ / [(1 – tan θ)] + sin θ / [(1 – cot θ)]
=cos θ / [1 – (sin θ/cos θ)] + sin θ/[1 – (cos θ/sin θ)]
= cos θ / [(cos θ – sin θ/cos θ] + sin θ / [(sin θ – cos θ/sin θ)]
= cos2θ/(cos θ – sin θ) + sin2θ/(cos θ – sin θ)
= (cos2θ – sin2θ)/(cos θ – sin θ)
= [(cos θ + sin θ)(cos θ – sin θ)] / (cos θ – sin θ)
= (cos θ + sin θ) = RHS
Hence proved.
Question 3: Prove, (tan θ + sec θ – 1)/(tan θ – sec θ + 1) = (1 + sin θ) × sec θ
Solution:
LHS = (tan θ + sec θ – 1)/(tan θ – sec θ + 1)
= [(tan θ + sec θ) – (sec2θ – tan2θ)]/(tan θ – sec θ + 1) [ sec2θ – tan2θ = 1]
= [ (tan θ + sec θ) – (sec θ + tan θ) (sec θ – tan θ)] / (tan θ – sec θ + 1)
= [ (tan θ + sec θ) × (1 – sec θ + tan θ)] / (tan θ – sec θ + 1)
= [(tan θ + sec θ) × (tan θ – sec θ + 1)] / (tan θ – sec θ + 1)
= (tan θ + sec θ)
= (sin θ/cos θ) + (1/cos θ)
= (sin θ + 1) / cos θ
= (1 + sin θ) × secθ = RHS ; Hence proved.
Question 4: = cosec θ – cot θ [Hint: Multiply numerator and denominator by (sec θ – 1)]
Solution:
LHS =
= (multiply numerator and denominator by (sec θ – 1))
=
=√[(sec θ -1)2 / tan2θ ] {sec2θ = 1 + tan2θ ⇢ sec2θ – 1 = tan2θ}
= (sec θ – 1) / tan θ
= (sec θ/tan θ) – (1/tan θ)
= [(1/cos θ) / (sin θ/cos θ)] – cot θ
= [(1/cos θ) × (cos θ/sin θ)] – cot θ
= (1/sin θ) – cot θ
= cosec θ – cot θ = RHS, Hence proved.
Question 5: Prove, (sin θ+cosec θ)2+(cos θ+sec θ)2=7+tan2(θ)+cot2(θ)
Solution:
LHS = {sin2θ + cosec2θ + 2 sinθ cosecθ } + {cos2θ+ sec2θ +2 cosθ secθ}
={sin2θ + cosec2θ + 2} + { cos2θ + sec2θ + 2}
=sin2θ + cos2θ + sec2θ + cosec2θ + 4
=sec2θ + cosec2θ + 5 [sin2θ +cos2θ = 1]
= (1+ tan2θ) + (1+ cot2θ) +5 [sec2θ = 1 + tan2θ ; cosec2θ = 1+ cot2θ ]
=7+ tan2(θ) + cot2(θ) = RHS
Hence proved.
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