I have the problems in trigonometric identities
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I have the problems in trigonometric identities
like prove that questions
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Step-by-step explanation:
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 - sin2 A)
= (1 - sin A)2/(cos2 A), [Since sin2 θ + cos2 θ = 1 ⇒ cos2 θ = 1 - sin2 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/(sec2 θ - 1)}
=√{(sec θ -1)2/tan2 θ}; [since, sec2 θ = 1 + tan2 θ ⇒ sec2 θ - 1 = tan2 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. tan4 θ + tan2 θ = sec4 θ - sec2 θ
Solution:
L.H.S = tan4 θ + tan2 θ
= tan2 θ (tan2 θ + 1)
= (sec2 θ - 1) (tan2 θ + 1) [since, tan2 θ = sec2 θ – 1]
= (sec2 θ - 1) sec2 θ [since, tan2 θ + 1 = sec2 θ]
= sec4 θ - sec2 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)}
= cos2 θ/(cos θ - sin θ) + sin2 θ/(cos θ - sin θ)
= (cos2 θ - sin2 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)/(csc2 A - cot2 A)
= (2 csc A)/1; [since, csc2 A = 1 + cot2 A ⇒ csc2A - cot2 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) - (sec2 θ - tan2 θ)]/(tan θ - sec θ + 1), [Since, sec2 θ - tan2 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.
Answer:
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