Trigonometry
- If cos α + sec α = √3 , then the value of cos3 α + sec3α is
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cos α + sec α = √3
∴ cos3 α + sec3 α = (cosα + secα)3 – 3 cosα . secα (cosα + secα)
= (√3)3 - 3 × √3
= 3√3 - 3√3 = 0Correct Option: C
cos α + sec α = √3
∴ cos3 α + sec3 α = (cosα + secα)3 – 3 cosα . secα (cosα + secα)
= (√3)3 - 3 × √3
= 3√3 - 3√3 = 0
- The value of θ (0 ≤ θ ≤ 90°) satisfying 2 sin²θ = 3 cos θ is
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2 sin²θ = 3 cosθ
⇒ 2(1 – cos²θ) = 3 cosθ
⇒ 2 – 2cos²θ = 3 cosθ
⇒ 2 cos²θ + 3cosθ – 2 = 0
⇒ 2 cos²θ + 4 cosθ – cos θ – 2 = 0
⇒ 2 cosθ (cosθ + 2) –1(cosθ + 2) = 0
⇒ (2cosθ – 1) (cosθ + 2) = 0
⇒ 2 cosθ – 1 = 0 because cosθ + 2 ≠ 0
⇒ 2 cosθ = 1⇒ cos θ = 1 = cos60° 2
⇒ θ = 60°Correct Option: A
2 sin²θ = 3 cosθ
⇒ 2(1 – cos²θ) = 3 cosθ
⇒ 2 – 2cos²θ = 3 cosθ
⇒ 2 cos²θ + 3cosθ – 2 = 0
⇒ 2 cos²θ + 4 cosθ – cos θ – 2 = 0
⇒ 2 cosθ (cosθ + 2) –1(cosθ + 2) = 0
⇒ (2cosθ – 1) (cosθ + 2) = 0
⇒ 2 cosθ – 1 = 0 because cosθ + 2 ≠ 0
⇒ 2 cosθ = 1⇒ cos θ = 1 = cos60° 2
⇒ θ = 60°
- If a (tanθ + cot θ ) = 1, sinθ + cos θ = b with 0° < θ < 90°, then a relation between a and b is
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a (tanθ + cotθ) = 1
⇒ a 
sin θ + cos θ 
= 1 cos θ sin θ ⇒ a sin² θ + cos² θ = 1 sin θ.cos θ
⇒ sinθ . cosθ = a ....(i)
sinθ + cosθ = b
On squaring both sides,
sin²θ + cos²θ + 2 sinθ . cosθ = b²
⇒ 1 + 2a = b²
⇒ 2a = b² –1Correct Option: C
a (tanθ + cotθ) = 1
⇒ a sin θ + cos θ = 1 cos θ sin θ ⇒ a sin² θ + cos² θ = 1 sin θ.cos θ
⇒ sinθ . cosθ = a ....(i)
sinθ + cosθ = b
On squaring both sides,
sin²θ + cos²θ + 2 sinθ . cosθ = b²
⇒ 1 + 2a = b²
⇒ 2a = b² –1
- If A is an acute angle and cot A + cosec A = 3, then the value of sin A is
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cosec²A – cot² A = 1
(cosec A + cot A) (cosec A – cot A) = 1cosec A – cot A = 1 3
cosec A + cot A = 3 On adding,2 cosec A = 1 + 3 = 1 + 9 = 10 3 3 3 ⇒ cosec A = 10 = 5 3 × 2 3 ∴ sin A = 3 5
Correct Option: B
cosec²A – cot² A = 1
(cosec A + cot A) (cosec A – cot A) = 1cosec A – cot A = 1 3
cosec A + cot A = 3 On adding,2 cosec A = 1 + 3 = 1 + 9 = 10 3 3 3 ⇒ cosec A = 10 = 5 3 × 2 3 ∴ sin A = 3 5
- The simplest value of sin²x + 2 tan²x – 2 sec²x + cos²x is
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sin² x + 2 tan² x – 2 sec² x + cos²x = sin²x + cos²x – 2 sec²x + 2 tan²x
= 1 – 2 (sec²x – tan²x)
= 1 – 2 = – 1
[sec²x – tan²x = 1, sin² x + cos² x = 1]Correct Option: A
sin² x + 2 tan² x – 2 sec² x + cos²x = sin²x + cos²x – 2 sec²x + 2 tan²x
= 1 – 2 (sec²x – tan²x)
= 1 – 2 = – 1
[sec²x – tan²x = 1, sin² x + cos² x = 1]
