Trigonometry
- What is the value of tan 56° ?
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tan 56° = tan(45° + 11°) = tan 45° + tan 11° 1 - tan45° tan11° ∵ tan (A + B) = tanA + tanB &becaose; tan 45° = 1 1 - tanA tanB = 1 + tan 11° 1 - tan 11° = 1 + sin 11° cos 11° 1 - sin 11° cos 11° = cos 11° + sin 11° cos 11° - sin 11° Correct Option: B
tan 56° = tan(45° + 11°) = tan 45° + tan 11° 1 - tan45° tan11° ∵ tan (A + B) = tanA + tanB &becaose; tan 45° = 1 1 - tanA tanB = 1 + tan 11° 1 - tan 11° = 1 + sin 11° cos 11° 1 - sin 11° cos 11° = cos 11° + sin 11° cos 11° - sin 11°
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The value of 3cosθ + cos3θ is equal to 3sinθ - sin3θ
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3cosθ + cos3θ 3sinθ - sin3θ = 3cosθ + 4 cos³θ - cos3θ 3sinθ - (3sinθ - 4sin³θ)
∵ cos3θ = 4cos³θ – 3cosθ
sin3θ = 3sinθ – 4sin³θ= 4cos³θ = cot³θ 4sin³θ Correct Option: B
3cosθ + cos3θ 3sinθ - sin3θ = 3cosθ + 4 cos³θ - cos3θ 3sinθ - (3sinθ - 4sin³θ)
∵ cos3θ = 4cos³θ – 3cosθ
sin3θ = 3sinθ – 4sin³θ= 4cos³θ = cot³θ 4sin³θ
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The value of sin 5π sin π is 12 12
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sin 5π .sin π 12 12 = 2 × 
2sin 5π sin π 
2 12 12 = 1 
cos 5π - π - cos 5π + π 
2 12 12 12 12
∵ 2sinA sinB = cos(A -B) - cos(A +B)= 1 cos 4π - cos 6π 2 12 12 = 1 cos π - cos π 2 3 2 = 1 × 1 2 2 = 1 4 Correct Option: A
sin 5π .sin π 12 12 = 2 × 2sin 5π sin π 2 12 12 = 1 cos 5π - π - cos 5π + π 2 12 12 12 12
∵ 2sinA sinB = cos(A -B) - cos(A +B)= 1 cos 4π - cos 6π 2 12 12 = 1 cos π - cos π 2 3 2 = 1 × 1 2 2 = 1 4
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The value of = cos(π + x) cos( -x) sin(π - x)cos 1 - cosB 2
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= cos(π + x) cos( -x) sin(π - x)cos 1 - cosB 2 = - cosx .cos x sinx(- sinx)
∵ cos(π + θ) = – cosθ
cos(– θ) = cosθ
sin(π – θ) = – sinθ= cos x × cos x sinx sinx
= cot x × cotx
= cot²x.Correct Option: C
= cos(π + x) cos( -x) sin(π - x)cos 1 - cosB 2 = - cosx .cos x sinx(- sinx)
∵ cos(π + θ) = – cosθ
cos(– θ) = cosθ
sin(π – θ) = – sinθ= cos x × cos x sinx sinx
= cot x × cotx
= cot²x.
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The value of = sin 50° + cos70° -2tan² 225° is equal to sin 130° cos 110°
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= sin 50° + cos70° -2tan² 225° sin 130° cos 110° = sin 50° + cos70° - 2[tan (180° - 45°)]² sin (180° - 50°) cos (180° - 70°) = sin 50° + cos 70° -2tan² 45° sin 50°) - cos 70°)
∵ sin(180° – θ) = sinθ
cos(180° – θ) = – cosθ
tan(180° + θ) = tanθ
= 1 – 1 – 2 = –2Correct Option: B
= sin 50° + cos70° -2tan² 225° sin 130° cos 110° = sin 50° + cos70° - 2[tan (180° - 45°)]² sin (180° - 50°) cos (180° - 70°) = sin 50° + cos 70° -2tan² 45° sin 50°) - cos 70°)
∵ sin(180° – θ) = sinθ
cos(180° – θ) = – cosθ
tan(180° + θ) = tanθ
= 1 – 1 – 2 = –2
