Trigonometry
- If tan 45° = cotθ, then the value of θ, in radians is
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tan 45° = cotθ
⇒ tan 45° = tan (90° – θ)
⇒ 45° = 90° – θ
⇒ θ = 90° – 45° = 45°
∵ 180° = π radian∴ 45° = π × 45° 180 = π radian 4 Correct Option: C
tan 45° = cotθ
⇒ tan 45° = tan (90° – θ)
⇒ 45° = 90° – θ
⇒ θ = 90° – 45° = 45°
∵ 180° = π radian∴ 45° = π × 45° 180 = π radian 4
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ABC is a triangle. If sin 
A + B 
= √3 , then the value of sin (C/2) is 2 2
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sin A + B = √3 = sin 60° 2 2 ⇒ A + B = 60° 2
⇒ A + B = 2 × 60° = 120°
∴ C = 180° – 120° = 60°∴ sin C = sin 30° = 1 2 2 Correct Option: C
sin A + B = √3 = sin 60° 2 2 ⇒ A + B = 60° 2
⇒ A + B = 2 × 60° = 120°
∴ C = 180° – 120° = 60°∴ sin C = sin 30° = 1 2 2
- Find the value of 8 cos 10° cos20° cos 40°.
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Expression = 8 cos 10°. cos 20° . cos 40°
Expression = 4 2 sin 10° .cos 10°. cos 20° . cos 40° sin 10° Expression = 2 2 sin 20° . cos 20° . cos 40° sin 10°
{ ∴ 2 sinθ . cosθ = sin2θ }Expression = 2 sin 40° . cos 40° sin 10° Expression = sin 80° = sin 80° sin 10° cos (90° - 10°) Expression = sin 80° or cos 10° cos 80° sin 10°
Expression = tan80° or cot 10°
{ ∴ cos (90° - θ) = sinθ and sin (90° - θ) = cosθ }Correct Option: C
Expression = 8 cos 10°. cos 20° . cos 40°
Expression = 4 2 sin 10° .cos 10°. cos 20° . cos 40° sin 10° Expression = 2 2 sin 20° . cos 20° . cos 40° sin 10°
{ ∴ 2 sinθ . cosθ = sin2θ }Expression = 2 sin 40° . cos 40° sin 10° Expression = sin 80° = sin 80° sin 10° cos (90° - 10°) Expression = sin 80° or cos 10° cos 80° sin 10°
Expression = tan80° or cot 10°
{ ∴ cos (90° - θ) = sinθ and sin (90° - θ) = cosθ }
- If secθ + tanθ = 2, then the value of sinθ is :
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secθ + tanθ = 2
∴ sec2θ – tan2θ = 1
⇒ (secθ + tanθ)(secθ – tanθ) = 1⇒ secθ – tanθ = 1 2 ∴ secθ + tanθ + secθ – tanθ = 2 + 1 2 ⇒ 2secθ = 5 ⇒ secθ = 5 2 4 Again, (secθ + tanθ) – (secθ – tanθ) = 2 - 1 2 ⇒ 2tanθ = 3 ⇒ tanθ = 3 2 4 ⇒ sinθ = tanθ = 3 ÷ 5 = 3 secθ 4 4 5 Correct Option: D
secθ + tanθ = 2
∴ sec2θ – tan2θ = 1
⇒ (secθ + tanθ)(secθ – tanθ) = 1⇒ secθ – tanθ = 1 2 ∴ secθ + tanθ + secθ – tanθ = 2 + 1 2 ⇒ 2secθ = 5 ⇒ secθ = 5 2 4 Again, (secθ + tanθ) – (secθ – tanθ) = 2 - 1 2 ⇒ 2tanθ = 3 ⇒ tanθ = 3 2 4 ⇒ sinθ = tanθ = 3 ÷ 5 = 3 secθ 4 4 5
- ∠Y is the right angle of the trianlge XYZ. If XY = 2 √6 cm and XZ – YZ = 2cm, then the value of (secX + tanX) is :
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XY = 2 6 cm
XY – YZ = 2 cm. ...(i)
∴ XZ2 = XY2 + YZ2
⇒ XZ2 - YZ2 = (2√6)2
⇒ XZ2 - YZ2 = 24∴ XZ2 - YZ2 = 24 XZ - YZ 2
⇒ XZ + YZ = 12 ....(ii)⇒ secX + tan X = XZ + YZ XY XY ⇒ secX + tan X = XZ + YZ = XY ⇒ secX + tan X = XZ + YZ = 12 = √6 XY 2√6
Correct Option: D
XY = 2 6 cm
XY – YZ = 2 cm. ...(i)
∴ XZ2 = XY2 + YZ2
⇒ XZ2 - YZ2 = (2√6)2
⇒ XZ2 - YZ2 = 24∴ XZ2 - YZ2 = 24 XZ - YZ 2
⇒ XZ + YZ = 12 ....(ii)⇒ secX + tan X = XZ + YZ XY XY ⇒ secX + tan X = XZ + YZ = XY ⇒ secX + tan X = XZ + YZ = 12 = √6 XY 2√6
