Trigonometry
- If 29 tan θ = 31, then the value of
1 + 2sin θ cos θ is equal to 1 - 2sin θ cos θ
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29 tanθ = 31 ⇒ tanθ = 31 29 Expression = 1 + 2sinθ .cosθ 1 - 2sinθ .cosθ = sin²θ + cos²θ + 2sin θ . cos θ sin²θ - cos²θ + 2sin θ . cos θ = (sinθ + cosθ)² (sinθ - cosθ)² 
= 
60 
² = (30)² = 900. 2
Correct Option: B
29 tanθ = 31 ⇒ tanθ = 31 29 Expression = 1 + 2sinθ .cosθ 1 - 2sinθ .cosθ = sin²θ + cos²θ + 2sin θ . cos θ sin²θ - cos²θ + 2sin θ . cos θ = (sinθ + cosθ)² (sinθ - cosθ)² = 60 ² = (30)² = 900. 2
- ABCD is a rectangle of which AC is a diagonal. The value of (tan² ∠CAD + 1) sin² ∠BAC is
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∠ACD = 45°
∠BAC = 45°
∴ (tan² ∠CAD + 1).sin² ∠BAC
= (tan² 45° + 1) sin² 45°=(1 + 1) × 1 = 2 × 1 = 1 √2 2
Correct Option: C
∠ACD = 45°
∠BAC = 45°
∴ (tan² ∠CAD + 1).sin² ∠BAC
= (tan² 45° + 1) sin² 45°=(1 + 1) × 1 = 2 × 1 = 1 √2 2
- If 0° < θ < 90° and 2 sin²θ + 3 cosθ = 3, then the value of θ is
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2 sin²θ + 3 cos θ = 3
⇒ 2 (1–cos²θ) + 3 cos θ = 3
⇒ 2 – 2 cos²θ + 3 cos θ = 3
⇒ 2 cos²θ – 3 cos θ + 1 = 0
⇒ 2 cos²θ – 2 cos θ – cos θ + 1 = 0
⇒ 2 cos θ (cos θ – 1) –1 (cos θ – 1) = 0
⇒ (2 cos θ – 1) (cos θ – 1) = 0
⇒ 2 cos θ – 1 = 0
⇒ 2 cos θ = 1⇒ cos θ = 1 ⇒ θ = 60° 2
or, cos θ – 1 = 0
⇒ cos θ = 1
⇒ θ = 0°Correct Option: B
2 sin²θ + 3 cos θ = 3
⇒ 2 (1–cos²θ) + 3 cos θ = 3
⇒ 2 – 2 cos²θ + 3 cos θ = 3
⇒ 2 cos²θ – 3 cos θ + 1 = 0
⇒ 2 cos²θ – 2 cos θ – cos θ + 1 = 0
⇒ 2 cos θ (cos θ – 1) –1 (cos θ – 1) = 0
⇒ (2 cos θ – 1) (cos θ – 1) = 0
⇒ 2 cos θ – 1 = 0
⇒ 2 cos θ = 1⇒ cos θ = 1 ⇒ θ = 60° 2
or, cos θ – 1 = 0
⇒ cos θ = 1
⇒ θ = 0°
- For any real values of θ,
-
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(Rationalising the numerator and the denominator)
= 1 - cosθ = 1 - cos θ sin θ sin θ sin θ
= cosecθ – cotθ.
Correct Option: C
(Rationalising the numerator and the denominator)= 1 - cosθ = 1 - cos θ sin θ sin θ sin θ
= cosecθ – cotθ.
- If the sum and difference of two angles are 135° and (π / 12) respectively, then the value of the angles in degree measure are
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Let the angles be A and B where A > B
∴ A + B = 135°
and, A – B= π = π × 180° = 15° 12 12 π
On adding
A + B + A – B
= 135° + 15° = 150°⇒ 2A = 150 ° ⇒ A = 150 = 75° 2
∴ A + B = 135°
⇒ B = 135° – 75° = 60°Correct Option: B
Let the angles be A and B where A > B
∴ A + B = 135°
and, A – B= π = π × 180° = 15° 12 12 π
On adding
A + B + A – B
= 135° + 15° = 150°⇒ 2A = 150 ° ⇒ A = 150 = 75° 2
∴ A + B = 135°
⇒ B = 135° – 75° = 60°
