Trigonometry
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If r sinθ = 7 and r cosθ = 7√3 , then the value of r is 2 2
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r sinθ = 7 ...(i) 2 r cosθ = 7√3 ...(ii) 2
On squaring both equations and adding
r²sin²θ + r² cos²θ = 47 + 147 4 4 ⇒ r² = 49 + 147 = 196 = 49 4 4
∴ r = √49 = 7
Correct Option: D
r sinθ = 7 ...(i) 2 r cosθ = 7√3 ...(ii) 2
On squaring both equations and addingr²sin²θ + r² cos²θ = 47 + 147 4 4 ⇒ r² = 49 + 147 = 196 = 49 4 4
∴ r = √49 = 7
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If θ + φ = π and sin θ = 1 , then the value of sin φ is 2 2
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sinθ = 1 = sin 30° = sin π 2 6 ⇒ θ = π 6
[∵ 180° = π radian]∴ θ + φ = π ⇒ π + φ = π 2 6 2 ⇒ φ = π - π = 3π - π 2 6 6 = 2π = π 6 3 ∴ sin φ = sin π = √3 3 2
Correct Option: D
sinθ = 1 = sin 30° = sin π 2 6 ⇒ θ = π 6
[∵ 180° = π radian]∴ θ + φ = π ⇒ π + φ = π 2 6 2 ⇒ φ = π - π = 3π - π 2 6 6 = 2π = π 6 3 ∴ sin φ = sin π = √3 3 2
- If tan x = sin 45°. cos 45° + sin 30° then the value of x is
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tanx = sin45°.cos45° + sin 30°
= 1 . 1 + 1 = 1 + 1 = 1 √2 √2 2 2 2
∴ tanx = tan 45° ⇒ x = 45°
Correct Option: B
tanx = sin45°.cos45° + sin 30°
= 1 . 1 + 1 = 1 + 1 = 1 √2 √2 2 2 2
∴ tanx = tan 45° ⇒ x = 45°
- If 3 sin θ + 5 cos θ = 5, then the value of 5 sin θ – 3 cos q will be
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3 sin θ + 5 cos θ = 5 --- (i)
5 sin θ – 3 cos θ = x (let)-- (ii)
On squaring and adding both the equations,
(3sin θ + 5 cos θ) 2 + (5 sin θ – 3 cos θ )² = 5² + x²
⇒ 9 sin²θ + 2² cos²θ + 30 sin θ .cos θ+ 25 sin² θ + 9 cos² θ – 30 sin θ . cos θ = 25 + x²
⇒ 9 sin²θ + 9 cos²θ + 25 cos2 θ + 25 sin²θ = 25 + x²
⇒ 9 (sin²θ + cos²θ) + 25 (cos²θ + sin²θ ) = 25 + x²
⇒ 9 + 25 = 25 + x²
⇒ x² = 9
⇒ x = ± 3Correct Option: A
3 sin θ + 5 cos θ = 5 --- (i)
5 sin θ – 3 cos θ = x (let)-- (ii)
On squaring and adding both the equations,
(3sin θ + 5 cos θ) 2 + (5 sin θ – 3 cos θ )² = 5² + x²
⇒ 9 sin²θ + 2² cos²θ + 30 sin θ .cos θ+ 25 sin² θ + 9 cos² θ – 30 sin θ . cos θ = 25 + x²
⇒ 9 sin²θ + 9 cos²θ + 25 cos2 θ + 25 sin²θ = 25 + x²
⇒ 9 (sin²θ + cos²θ) + 25 (cos²θ + sin²θ ) = 25 + x²
⇒ 9 + 25 = 25 + x²
⇒ x² = 9
⇒ x = ± 3
- If θ is an acute angle and tan θ + cot θ = 2, then the value of tan5 θ + cot5 θ is
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tan θ + cot θ = 2
⇒ tan θ + 1 = 2 tan θ ⇒ tan² θ + 1 = 2 tan θ
⇒ tan² θ +1 = 2 tanθ
⇒ tan²θ – 2 tan θ + 1 = 0
⇒ (tan θ –1)² = 0
⇒ tanθ – 1= 0 ⇒ tanθ = 1∴ cot θ = 1 = 1 tan θ
∴ tan5θ + cot5θ = 1 + 1 = 2Correct Option: B
tan θ + cot θ = 2
⇒ tan θ + 1 = 2 tan θ ⇒ tan² θ + 1 = 2 tan θ
⇒ tan² θ +1 = 2 tanθ
⇒ tan²θ – 2 tan θ + 1 = 0
⇒ (tan θ –1)² = 0
⇒ tanθ – 1= 0 ⇒ tanθ = 1∴ cot θ = 1 = 1 tan θ
∴ tan5θ + cot5θ = 1 + 1 = 2
