Trigonometrical Identity
Definition of trigonometrical identity:
An equation which is true for all values of the variable
involved is called an identity. An equation which involves trigonometric ratios
of an angle and is true for all the values of the angle is called trigonometrical
identities.
When the solutions of any trigonometric ratio problems represent the same
expression in the L.H.S. and R.H.S. and the relation is satisfied for all the values
of θ then such relation is
called a trigonometrical identity.
Mutual relations among the trigonometrical ratios are generally used to
establish the equality of such trigonometrical identities.
To solve different types of trignometrical identity follow the formula:
● sin θ ∙ csc θ = 1 ⇒ csc θ = 1/sin θ
● cos θ ∙ sec θ = 1 ⇒ sec θ = 1/cos θ
● tan θ ∙ cot θ = 1 ⇒ cot θ = 1/tan θ
● tan θ = sin θ/cos θ
● cot θ = cos θ/sin θ
● sin
^{2} θ implies (sin θ)
^{2} similarly, tan
^{3} θ means (tan θ)
^{3} etc.
● sin
^{2} θ + cos
^{2} θ = 1
cos
^{2} θ = 1 - sin
^{2} θ
sin
^{2} θ = 1 - cos
^{2} θ
● sec
^{2} θ = 1 + tan
^{2} θ
sec
^{2} θ - tan
^{2} θ = 1
tan
^{2} θ = sec
^{2} θ - 1
● csc
^{2} θ = 1 + cot
^{2} θ
csc
^{2} θ - 1 = cot
^{2} θ
csc
^{2} θ - cot
^{2} θ = 1
● The trigonometrical ratios of a positive acute angle θ are always non-negative and
(i) sin θ and cos θ can never be greater than 1;
(ii) sec θ and csc θ can never be less than 1;
(iii) tan θ and cot θ can have any value.
Worked-out
problems on trigonometric identity:
1. Proof the
identity:
tan
^{2} θ – (1/cos
^{2} θ) + 1 = 0
Solution:
L.H.S = tan
^{2} θ – (1/cos
^{2} θ) + 1
= tan
^{2} θ - sec
^{2} θ + 1 [since, 1/cos θ = sec θ]
= tan
^{2} θ – (1 + tan
^{2} θ) +1 [since, sec
^{2} θ = 1 + tan
^{2} θ]
= tan
^{2} θ – 1 – tan
^{2} θ + 1
= 0 = R.H.S.
Proved
2. Verify that:
1/(sin θ + cos θ) + 1/(sin θ - cos θ) = 2 sin θ/(1 – 2 cos
^{2} θ)
Solution:
L.H.S = 1/(sin θ + cos θ) + 1/(sin θ - cos θ)
= [(sin θ - cos θ) + (sin θ + cos θ)]/(sin θ + cos θ)(sin θ - cos θ)
= [sin θ - cos θ + sin θ + cos θ]/(sin
^{2} θ - cos
^{2} θ)
= 2 sin θ/[(1 - cos
^{2} θ) - cos
^{2} θ] [since, sin
^{2} θ = 1 - cos
^{2} θ]
= 2 sin θ/[1 - cos
^{2} θ - cos
^{2} θ]
= 2 sin θ/[1 – 2 cos
^{2} θ] = R.H.S.
Proved
3. Prove that:
sec
^{2} θ + csc
^{2} θ = sec
^{2} θ ∙ csc
^{2} θ
Solution:
L.H.S. = sec
^{2} θ + csc
^{2} θ
= 1/cos
^{2} θ + 1/sin
^{2} θ [since, sec θ = 1/cos θ and csc θ = 1/sin θ]
= (sin
^{2} θ + cos
^{2} θ)/(cos
^{2} θ sin
^{2} θ)
= 1/cos
^{2} θ ∙ sin
^{2} θ [since, sin
^{2} θ + cos
^{2} θ = 1]
= 1/cos
^{2} θ ∙ 1/sin
^{2} θ
= sec
^{2} θ ∙ csc
^{2} θ = R.H.S.
Proved
More examples on trigonometrical identity are explained below. To proof the identities step-by-step follow the above trig formulas.
4. Prove the identity:
cos θ/(1 + sin θ) = (1 + cos θ - sin θ)/(1 + cos θ + sin θ)
Solution:
R. H. S. = (1 + cos θ - sin θ)/(1 + cos θ + sin θ)
= {(1 + cos θ - sin θ) (1 + cos θ + sin θ)}/{(1+ cos θ + sin θ) (1 + cos θ + sin θ)} [multiplying both numerator and denominator by (1 + cos θ + sin θ)]
= {(1 + cos θ)
^{2} - sin
^{2} θ}/(1 + cos θ + sin θ)
^{2}
= (1 + cos
^{2} θ + 2 cos θ - sin
^{2} θ)/{(1 + cos θ)
^{2} + 2 ∙ (1 + cos θ) sin θ + sin
^{2} θ}
= (cos
^{2} θ + 2 cos θ + 1 - sin
^{2} θ)/{1 + cos
^{2} θ + 2 cos θ + 2 ∙ (1 + cos θ) ∙ sin θ + sin
^{2} θ}
= (cos
^{2} θ + 2 cos θ + cos
^{2} θ)/{2 + 2 cos θ + 2 ∙ (1 + cos θ) ∙ sin θ} [since, sin
^{2} θ + cos
^{2} θ = 1 and 1 - sin
^{2} θ = cos
^{2} θ]
= {2 cos θ (1 + cos θ)}/{2 (1 + cos θ)(1 + sin θ)}
= cos θ/(1 + sin θ) = L.H.S.
Proved
5. Verify the trigonometrical identity:
(cot θ + csc θ – 1)/(cot θ - csc θ + 1) = (1 + cos θ)/sin θ
L.H.S. = (cot θ + csc θ – 1)/(cot θ - csc θ + 1)
= {cot θ + csc θ - (csc
^{2} θ - cot
^{2} θ)}/(cot θ - csc θ + 1)
[csc
^{2} θ = 1 + cot
^{2} θ ⇒ csc
^{2} θ - cot
^{2} θ = 1]
= {(cot θ + csc θ) - (csc θ + cot θ) (csc θ - cot θ)}/(cot θ - csc θ + 1)
= {(cot θ + csc θ) (1 - csc θ + cot θ)}/ (1 - csc θ + cot θ)
= cot θ + csc θ
= (cos θ/sin θ) + (1/sin θ)
= (1 + cos θ)/sin θ = R.H.S.
Proved
● Trigonometric Functions
10th Grade Math
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