Trigonometry
- The value of sin² 65° + sin² 25° + cos² 35° + cos² 55° is
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Expression = sin² 65° + sin² 25° + cos²35° + cos² 55°
= sin² 65° + sin² (90° – 65°) + cos² 35° + cos² (90° – 35°)
= sin² 65° + cos² 65° + cos² 35° + sin² 35°
= 1 + 1 = 2Correct Option: C
Expression = sin² 65° + sin² 25° + cos²35° + cos² 55°
= sin² 65° + sin² (90° – 65°) + cos² 35° + cos² (90° – 35°)
= sin² 65° + cos² 65° + cos² 35° + sin² 35°
= 1 + 1 = 2
- If x sin 60°.tan 30° = sec 60°.cot 45°, then the value of x is
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x.sin 60°.tan 30°
= sec 60°.cot 45°⇒ x × √3 × 1 = 2 × 1 2 √3
⇒ x = 2 × 2 = 4Correct Option: C
x.sin 60°.tan 30°
= sec 60°.cot 45°⇒ x × √3 × 1 = 2 × 1 2 √3
⇒ x = 2 × 2 = 4
- Let A, B, C, D be the angles of a quadrilateral. If they are concyclic, then the value of cos A + cos B + cos C + cos D is
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ABCD is a concyclic quadrilateral.
∠A + ∠C = ∠B + ∠D = 180°
∴ ∠A = 180° – ∠C
∴ cos A = cos (180° – C) = – cos C
and cos B = – cos D
∴ cos A + cos B + cos C + cos D
= cos A + cos B – cos A – cos B = 0Correct Option: A
ABCD is a concyclic quadrilateral.
∠A + ∠C = ∠B + ∠D = 180°
∴ ∠A = 180° – ∠C
∴ cos A = cos (180° – C) = – cos C
and cos B = – cos D
∴ cos A + cos B + cos C + cos D
= cos A + cos B – cos A – cos B = 0
- If 0° < A < 90°, then the value of tan²A + cot² A – sec² A cosec² A is
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tan²A + cot²A – sec²A . cosec²A
= tan²A + cot²A – (1 + tan²A) (1 + cot²A)
= tan²A + cot²A – (1 + tan²A +
cot²A + cot²A.tan²A)
= tan²A + cot²A – 1 – tan²A – cot²A – cot²A . tan²A
= – 1 – 1 = – 2
[ tanA . cotA = 1]Correct Option: D
tan²A + cot²A – sec²A . cosec²A
= tan²A + cot²A – (1 + tan²A) (1 + cot²A)
= tan²A + cot²A – (1 + tan²A +
cot²A + cot²A.tan²A)
= tan²A + cot²A – 1 – tan²A – cot²A – cot²A . tan²A
= – 1 – 1 = – 2
[ tanA . cotA = 1]
- If α and β are positive acute angles, sin (4α – β) = 1 and cos (2 α + β) = (1 / 2) , then the value of sin (α + 2β) is
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sin (4α – β) = 1 = sin 90°
⇒ 4α – β = 90° ...(i)cos (2α + β) = 1 = cos 60° 2
⇒ 2α + β = 60° ...(ii)
On adding equations (i) and (ii),
4α - β + 2α + β = 90° + 60°⇒ 6α = 150° ⇒ α = 150 = 25° 6
From equation (ii),
2 × 25 + β = 60°
⇒ β = 60° – 50° = 10°
∴ sin (α + 2β)
= sin (25 + 2 × 10)= sin 45° = 1 = 25° √2
Correct Option: D
sin (4α – β) = 1 = sin 90°
⇒ 4α – β = 90° ...(i)cos (2α + β) = 1 = cos 60° 2
⇒ 2α + β = 60° ...(ii)
On adding equations (i) and (ii),
4α - β + 2α + β = 90° + 60°⇒ 6α = 150° ⇒ α = 150 = 25° 6
From equation (ii),
2 × 25 + β = 60°
⇒ β = 60° – 50° = 10°
∴ sin (α + 2β)
= sin (25 + 2 × 10)= sin 45° = 1 = 25° √2
