Trigonometry
- Which of the folliwing is correct ?
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We know that,
1R = 
180° 
π
1R = 57.29°
As we know that the value of sinq increases as &thetea; increases.
⇒ sin 1° < sin 57.29°sin 1° < sin 180 ° π
⇒ sin 1° < sin 1Correct Option: B
We know that,
1R = 180° π
1R = 57.29°
As we know that the value of sinq increases as &thetea; increases.
⇒ sin 1° < sin 57.29°sin 1° < sin 180 ° π
⇒ sin 1° < sin 1
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If tanα = m , tanβ = 1 then α + β equal to m + n 2m + n
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We know that,
tan(α + β) = tanα + tanβ 1 - tanα . tanβ
⇒ tan(α + β)= m + 1 m + 1 2m + 1 1 - m 1 (m + 1) (2m + 1) ∵ tan α = m m + 1 ∵ tan β = m 2m + 1 = 2m² + m + m + 1 (m + 1)(2m + 1) 2m² + 3m + 1 - m (m + 1)(2m + 1) = 2m² + 2m + 1 = 1 2m² + 2m + 1
⇒ tan(α + β) = 1tan(α + β) = tan π 4 ∴ α + β = π 4 Correct Option: D
We know that,
tan(α + β) = tanα + tanβ 1 - tanα . tanβ
⇒ tan(α + β)= m + 1 m + 1 2m + 1 1 - m 1 (m + 1) (2m + 1) ∵ tan α = m m + 1 ∵ tan β = m 2m + 1 = 2m² + m + m + 1 (m + 1)(2m + 1) 2m² + 3m + 1 - m (m + 1)(2m + 1) = 2m² + 2m + 1 = 1 2m² + 2m + 1
⇒ tan(α + β) = 1tan(α + β) = tan π 4 ∴ α + β = π 4
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If α + β = π , then the value of (1 + tanα)(1 + tanβ) is 4
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Here, α + β = π 4
(1 + tanα)(1 + tanβ)
= 1 + tanβ + tanα + tanα tanβ
= 1 + tanα + tanβ + tanα tanβ
Also, we know that,tan (α + β) = tanα + tanβ 1 - tanαtanβ tan π = tanα + tanβ 4 1 - tanαtanβ
⇒ 1 - tanαtanβ = tanα + tanβ
⇒ (1 + tanα)(1 + tanβ)
= 1 + 1 – tanα tanβ + tanα tanβ = 2Correct Option: B
Here, α + β = π 4
(1 + tanα)(1 + tanβ)
= 1 + tanβ + tanα + tanα tanβ
= 1 + tanα + tanβ + tanα tanβ
Also, we know that,tan (α + β) = tanα + tanβ 1 - tanαtanβ tan π = tanα + tanβ 4 1 - tanαtanβ
⇒ 1 - tanαtanβ = tanα + tanβ
⇒ (1 + tanα)(1 + tanβ)
= 1 + 1 – tanα tanβ + tanα tanβ = 2
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If tan A = 1 - cos B , then tan 2A is equal to sin B
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Here, tanA = 1 - cosB sinB
We know that,tan2A = 2tanA 1 - tan²A tan2A = 2 1 - cosB sinB 1 - 1 - cosB ² sinB tan2A = 2(1 - cosB) sinB sin²B - (1 - cosB)² sin²B = 2(1 - cosB)sinB [∵ sin²θ = 1 - cos²θ] 1 - cos²B - (1 - cosB)² = 2(1 - cosB)sinB (1 - cosB)[1 + cosB - 1 + cosB] = 2sinB 2cosB
= tan B.Correct Option: B
Here, tanA = 1 - cosB sinB
We know that,tan2A = 2tanA 1 - tan²A tan2A = 2 1 - cosB sinB 1 - 1 - cosB ² sinB tan2A = 2(1 - cosB) sinB sin²B - (1 - cosB)² sin²B = 2(1 - cosB)sinB [∵ sin²θ = 1 - cos²θ] 1 - cos²B - (1 - cosB)² = 2(1 - cosB)sinB (1 - cosB)[1 + cosB - 1 + cosB] = 2sinB 2cosB
= tan B.
- The value of sin (45° + θ) – cos(45° – θ) is
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sin (45° + θ) – cos (45° – θ)
= sin 45° cosθ + cos 45° sinθ – (cos 45° cosθ + sin 45° sinθ)
∵ sin (A + B) = sinA cosB + cosA sinB
cos(A – B) = cosA cosB + sinA sinB= cosθ + sinθ - cosθ - sinθ = 0 √2 √2 √2 √2 Correct Option: D
sin (45° + θ) – cos (45° – θ)
= sin 45° cosθ + cos 45° sinθ – (cos 45° cosθ + sin 45° sinθ)
∵ sin (A + B) = sinA cosB + cosA sinB
cos(A – B) = cosA cosB + sinA sinB= cosθ + sinθ - cosθ - sinθ = 0 √2 √2 √2 √2
