Problems on Trigonometric Identities
Here we
will prove the problems on trigonometric identities. In an identity there are
two sides of the equation, one side is known as ‘left hand side’ and the other
side is known as ‘right hand side’ and to prove the identity we need to use
logical steps showing that one side of the equation ends up with the other side
of the equation.
Proving the problems on trigonometric
identities:
1. (1 - sin A)/(1 + sin A) = (sec A - tan A)
^{2}
Solution:
L.H.S = (1 - sin A)/(1 + sin A)
= (1 - sin A)
^{2}/(1 - sin A) (1 + sin A),[Multiply both numerator and denominator by (1 - sin A)
= (1 - sin A)
^{2}/(1 - sin
^{2} A)
= (1 - sin A)
^{2}/(cos
^{2} A), [Since sin
^{2} θ + cos
^{2} θ = 1 ⇒ cos
^{2} θ = 1 - sin
^{2} θ]
= {(1 - sin A)/cos A}
^{2}
= (1/cos A - sin A/cos A)
^{2}
= (sec A – tan A)
^{2} = R.H.S.
Proved.
2. Prove that, √{(sec θ – 1)/(sec θ + 1)} = cosec θ - cot θ.
Solution:
L.H.S.= √{(sec θ – 1)/(sec θ + 1)}
= √[{(sec θ - 1) (sec θ - 1)}/{(sec θ + 1) (sec θ - 1)}]; [multiplying numerator and denominator by (sec θ - l) under radical sign]
= √{(sec θ - 1)
^{2}/(sec
^{2} θ - 1)}
=√{(sec θ -1)
^{2}/tan
^{2} θ}; [since, sec
^{2} θ = 1 + tan
^{2} θ ⇒ sec
^{2} θ - 1 = tan
^{2} θ]
= (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 = R.H.S.
Proved.
3. tan
^{4} θ + tan
^{2} θ = sec
^{4} θ - sec
^{2} θ
Solution:
L.H.S = tan
^{4} θ + tan
^{2} θ
= tan
^{2} θ (tan
^{2} θ + 1)
= (sec
^{2} θ - 1) (tan
^{2} θ + 1) [since, tan
^{2} θ = sec
^{2} θ – 1]
= (sec
^{2} θ - 1) sec
^{2} θ [since, tan
^{2} θ + 1 = sec
^{2} θ]
= sec
^{4} θ - sec
^{2} θ = R.H.S.
Proved.
More problems on trigonometric identities are shown where one side of the identity ends up with the other side.
4. . cos θ/(1 - tan θ) + sin θ/(1 - cot θ) = sin θ + cos θ
Solution:
L.H.S = 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)}
= cos
^{2} θ/(cos θ - sin θ) + sin
^{2} θ/(cos θ - sin θ)
= (cos
^{2} θ - sin
^{2} θ)/(cos θ - sin θ)
= [(cos θ + sin θ)(cos θ - sin θ)]/(cos θ - sin θ)
= (cos θ + sin θ) = R.H.S.
Proved.
5. Show that, 1/(csc A - cot A) - 1/sin A = 1/sin A - 1/(csc A + cot A)
Solution:
We have,
1/(csc A - cot A) + 1/(csc A + cot A)
= (csc A + cot A + csc A - cot A)/(csc
^{2} A - cot
^{2} A)
= (2 csc A)/1; [since, csc
^{2} A = 1 + cot
^{2} A ⇒ csc
^{2}A - cot
^{2} A = 1]
= 2/sin A; [since, csc A = 1/sin A]
Therefore,
1/(csc A - cot A) + 1/(csc A + cot A) = 2/sin A
⇒ 1/(csc A - cot A) + 1/(csc A + cot A) = 1/sin A + 1/sin A
Therefore, 1/(csc A - cot A) - 1/sin A = 1/sin A - 1/(csc A + cot A)
Proved.
6. (tan θ + sec θ - 1)/(tan θ - sec θ + 1) = (1 + sin θ)/cos θ
Solution:
L.H.S = (tan θ + sec θ - 1)/(tan θ - sec θ + 1)
= [(tan θ + sec θ) - (sec
^{2} θ - tan
^{2} θ)]/(tan θ - sec θ + 1), [Since, sec
^{2} θ - tan
^{2} θ = 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)/cos θ = R.H.S.
Proved.
● Trigonometric Functions
10th Grade Math
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