Trigonometry
- In a ∆ ABC, ∠B = (π / 3) , ∠C = (π / 4) and D divides BC internally in the ratio 1 : 3 then sin (sin ∠BAD / sin∠CAD) is equal to
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∠B = π , ∠ = π 3 4 and BD = 1 DC 3
From ∆ ABD,BD = AD sin BAD sin ABD ⇒ BD = AD sin BAD sin (π / 3) ⇒ BD = AD sin BAD (√3 / 2) ⇒ AD = √3 . BD ...(i) 2 sin BAD
From ∆ ADC,CD = AD sin DAC sin ACD ⇒ CD = AD sin DAC sin (π / 4) ⇒ AD = 1 . CD ...(ii) √2 sin DAC
From equations (i) and (ii),√3 . BD = 1 . CD 2 sin BAD √2 sin DAC 
= sin BAD - √3 × √2 × 1 sin DAC 2 3 = 1 = 1 √2 × √3 √6
Correct Option: C
∠B = π , ∠ = π 3 4 and BD = 1 DC 3
From ∆ ABD,BD = AD sin BAD sin ABD ⇒ BD = AD sin BAD sin (π / 3) ⇒ BD = AD sin BAD (√3 / 2) ⇒ AD = √3 . BD ...(i) 2 sin BAD
From ∆ ADC,CD = AD sin DAC sin ACD ⇒ CD = AD sin DAC sin (π / 4) ⇒ AD = 1 . CD ...(ii) √2 sin DAC
From equations (i) and (ii),√3 . BD = 1 . CD 2 sin BAD √2 sin DAC = sin BAD - √3 × √2 × 1 sin DAC 2 3 = 1 = 1 √2 × √3 √6
- If sin 3A = cos ( A – 26°), where 3A is an acute angle then the value of A is
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sin 3A = cos (A – 26°)
⇒ cos (90° – 3A) = cos (A – 26°)
⇒ 90° – 3A = A – 26°
⇒ 90° + 26° = 3A + A
⇒ 4A = 116⇒A = 116 = 29° 4
Correct Option: A
sin 3A = cos (A – 26°)
⇒ cos (90° – 3A) = cos (A – 26°)
⇒ 90° – 3A = A – 26°
⇒ 90° + 26° = 3A + A
⇒ 4A = 116⇒A = 116 = 29° 4
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Value of sec² θ - sin² θ - 2sin 4 θ is 2cos4θ - cos² θ
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sec²θ - sin²θ - 2sin4θ 2cos4θ - cos²θ sec²θ - sin²θ (1 - 2sin²θ) cos²θ(2cos²θ - 1) sec²θ - sin²θ (1 - 2(1 - 2sin²θ)) cos²θ(2cos²θ - 1) sec²θ - tan²θ (2 cos²θ - 1) 2 cos²θ - 1
= sec²θ - tan²θ = 1Correct Option: A
sec²θ - sin²θ - 2sin4θ 2cos4θ - cos²θ sec²θ - sin²θ (1 - 2sin²θ) cos²θ(2cos²θ - 1) sec²θ - sin²θ (1 - 2(1 - 2sin²θ)) cos²θ(2cos²θ - 1) sec²θ - tan²θ (2 cos²θ - 1) 2 cos²θ - 1
= sec²θ - tan²θ = 1
- If x = a (sinθ + cosθ), y = b (sinθ – cosθ) then the value of
x² + y² is a² b²
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x = a (sinθ + cosθ) and
y = b (sinθ – cosθ)⇒ x = sin θ + cos θ and a y = sin θ - cos θ and b ∴ x² + y² = (sinθ + cosθ)² +(sinθ – cosθ)
2a² b²
= sin²θ + cos²θ + 2 sinθ.cosθ +
sin²θ + cos²θ – 2sinθ . cosθ
= 2 (sin²θ + cos²θ) = 2Correct Option: C
x = a (sinθ + cosθ) and
y = b (sinθ – cosθ)⇒ x = sin θ + cos θ and a y = sin θ - cos θ and b ∴ x² + y² = (sinθ + cosθ)² +(sinθ – cosθ)
2a² b²
= sin²θ + cos²θ + 2 sinθ.cosθ +
sin²θ + cos²θ – 2sinθ . cosθ
= 2 (sin²θ + cos²θ) = 2
- If sin 5 θ = cos 20° ( 0° < θ < 90°) then the value of θ is
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sin 5θ = cos20°
⇒ sin 5θ = sin (90 – 20) = sin 70°
⇒ 5θ = 70°⇒ θ = 70 = 14° 5
Correct Option: D
sin 5θ = cos20°
⇒ sin 5θ = sin (90 – 20) = sin 70°
⇒ 5θ = 70°⇒ θ = 70 = 14° 5
