Trigonometry
- If a sin θ + b cos θ = c, then a cos θ – b sin θ is equal to
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a sinθ + b cosθ = c .... (i)
a cosθ – b sinθ = x (let) ... (ii)
On squaring equations (i) and (ii) and adding, a 2sin²θ + b 2cos²θ + 2ab sinθ. cosθ + a 2cos²θ + b 2sin²θ – 2absinθ. cosθ = c² + x²
⇒ a² (sin2θ + cos2θ) + b 2 (cos²θ + sin²θ) = c² + x²
⇒ a² + b² = c²² + x²
⇒ x² = a² + b² – c²
⇒ x = ± √a² + b² -c²Correct Option: C
a sinθ + b cosθ = c .... (i)
a cosθ – b sinθ = x (let) ... (ii)
On squaring equations (i) and (ii) and adding, a 2sin²θ + b 2cos²θ + 2ab sinθ. cosθ + a 2cos²θ + b 2sin²θ – 2absinθ. cosθ = c² + x²
⇒ a² (sin2θ + cos2θ) + b 2 (cos²θ + sin²θ) = c² + x²
⇒ a² + b² = c²² + x²
⇒ x² = a² + b² – c²
⇒ x = ± √a² + b² -c²
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If tan α = 2, then the value of cosec²α - sec²α is cosec²α + sec²α
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tan α = 2
∴ cosec²α - sec²α cosec²α + sec²α = 1 + cot²α - 1 - tan²α cot²α + tan²α + 2 = cot²α - tan²α cot²α + tan²α + 2 
= - 15 = - 3 25 5
Correct Option: B
tan α = 2
∴ cosec²α - sec²α cosec²α + sec²α = 1 + cot²α - 1 - tan²α cot²α + tan²α + 2 = cot²α - tan²α cot²α + tan²α + 2 = - 15 = - 3 25 5
- The value of (1 + sec 20° + cot 70°) (1 – cosec 20° + tan 70°) is equal to
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(1 + sec 20° + cot 70°) (1 – cosec 20° + tan 70°)
= (1 + sec 20° + tan 20°) (1 – cosec 20° + cot 20°)
[∵ tan (90° – θ) = cotθ ; cot (90° – θ) = tanθ]= 
1 + 1 + sin 20° 
cos 20° cos 20° = 1 - 1 + cos 20° sin 20° sin 20° = cos 20° + 1 + sin 20° . sin 20° - 1 + cos 20° cos 20° sin 20° = (cos 20° + sin 20°)² - 1 sin 20° . cos 20° = cos² 20° + sin² 20° + 2 sin 20° . cos 20° - 1 sin 20° . cos 20° = 1 + sin 20° . cos 20° - 1 = 2 sin 20° . cos 20°
Correct Option: C
(1 + sec 20° + cot 70°) (1 – cosec 20° + tan 70°)
= (1 + sec 20° + tan 20°) (1 – cosec 20° + cot 20°)
[∵ tan (90° – θ) = cotθ ; cot (90° – θ) = tanθ]= 1 + 1 + sin 20° cos 20° cos 20° = 1 - 1 + cos 20° sin 20° sin 20° = cos 20° + 1 + sin 20° . sin 20° - 1 + cos 20° cos 20° sin 20° = (cos 20° + sin 20°)² - 1 sin 20° . cos 20° = cos² 20° + sin² 20° + 2 sin 20° . cos 20° - 1 sin 20° . cos 20° = 1 + sin 20° . cos 20° - 1 = 2 sin 20° . cos 20°
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If 0° < A < 90°, the value of tan A – sec A –1 is tan A + sec A + 1
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= tanA - secA - 1 tanA + secA + 1 = tanA - secA - (sec²A - tan²A) tanA + secA + 1 = (tanA – secA) – (secA – tanA)(secA tanA) tanA + secA + 1 = (tanA – secA) + (tanA – secA)(secA tanA) tanA + secA + 1 = (tanA – secA)(1 secA tanA) tanA + secA + 1 = tanA – secA = sinA - 1 cosA cosA = sinA - 1 cosA
Correct Option: A
= tanA - secA - 1 tanA + secA + 1 = tanA - secA - (sec²A - tan²A) tanA + secA + 1 = (tanA – secA) – (secA – tanA)(secA tanA) tanA + secA + 1 = (tanA – secA) + (tanA – secA)(secA tanA) tanA + secA + 1 = (tanA – secA)(1 secA tanA) tanA + secA + 1 = tanA – secA = sinA - 1 cosA cosA = sinA - 1 cosA
- If α is an cute angle and 2 sin α + 15 cos²α = 7 then the value of cot α is
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2 sinα + 15 cos²α = 7
⇒ 2 sinα + 15 (1 – sin²α) = 7
⇒ 2 sinα + 15 – 15 sin²α = 7
⇒ 15 sin²α – 2 sinα – 15 + 7 = 0
⇒ 15 sin²α – 2 sinα – 8 = 0
⇒ 15 sin²α – 12 sinα + 10 sinα – 8 = 0
⇒ 3 sinα (5 sinα – 4) + 2 (5 sinα – 4) = 0
⇒ (3 sinα + 2) (5 sinα – 4) = 0⇒ 5 sinα – 4 = 0 ⇒ sinα = 4 5 sinα ≠ - 2 because α is acute angle.
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∴ cotα = √cosec²α - 1
Correct Option: D
2 sinα + 15 cos²α = 7
⇒ 2 sinα + 15 (1 – sin²α) = 7
⇒ 2 sinα + 15 – 15 sin²α = 7
⇒ 15 sin²α – 2 sinα – 15 + 7 = 0
⇒ 15 sin²α – 2 sinα – 8 = 0
⇒ 15 sin²α – 12 sinα + 10 sinα – 8 = 0
⇒ 3 sinα (5 sinα – 4) + 2 (5 sinα – 4) = 0
⇒ (3 sinα + 2) (5 sinα – 4) = 0⇒ 5 sinα – 4 = 0 ⇒ sinα = 4 5 sinα ≠ - 2 because α is acute angle.
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∴ cotα = √cosec²α - 1
