Trigonometry
- If A, B and C be the angles of a triangle, then put of the following, the incorrect relation is :
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A + B + C = π
⇒ A + B = π - C 2 2 2 ⇒ sin 
A + B 
2 = sin π - C 2 2 = cos C 2
Similarly,cos A + B = sin C 2 2 cot A + B = tan C 2 2 tan A + B = cot C 2 2
Correct Option: C
A + B + C = π
⇒ A + B = π - C 2 2 2 ⇒ sin A + B 2 = sin π - C 2 2 = cos C 2
Similarly,cos A + B = sin C 2 2 cot A + B = tan C 2 2 tan A + B = cot C 2 2
- If 0 < x < (π / 2) and secx = cosecy, then the value of sin (x + y) is :
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secx = cosecy
⇒ cosx = siny⇒ sin π - x = siny 2 ⇒ y = π - x 2 ⇒ x + y = π 2 ∴ sin (x + y) = sin π = 1 2
Correct Option: B
secx = cosecy
⇒ cosx = siny⇒ sin π - x = siny 2 ⇒ y = π - x 2 ⇒ x + y = π 2 ∴ sin (x + y) = sin π = 1 2
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If sec²θ tan²θ= 7 , then sec4θ - tan4θ = 12
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sec⊃ θ – tan⊃ θ = 1
sec⊃ θ + tan⊃ θ = 7 12
∴ sec4θ – tan4θ
= (sec⊃ θ – tan⊃ θ) (sec⊃ θ + tan⊃ θ)= 1 × 7 = 7 12 12
Correct Option: A
sec⊃ θ – tan⊃ θ = 1
sec⊃ θ + tan⊃ θ = 7 12
∴ sec4θ – tan4θ
= (sec⊃ θ – tan⊃ θ) (sec⊃ θ + tan⊃ θ)= 1 × 7 = 7 12 12
- If θ be acute angle and cos θ = (15 / 17) , then the value of cot
(90° – θ) is
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cos θ = 15 7 ⇒ sec θ = 1 = 17 cos θ 15
∴ cot (90° – θ) = tan θ
√sec⊃ θ - 1
Correct Option: B
cos θ = 15 7 ⇒ sec θ = 1 = 17 cos θ 15
∴ cot (90° – θ) = tan θ
√sec⊃ θ - 1
- If tan (2θ + 45°) = cot 3θ where (2θ + 45°) and 3θ are acute angles, then the value of θ is
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tan (2θ + 45°) = cot 3θ
= tan (90° – 3θ )
⇒ 2θ + 45° = 90° – 3θ
⇒ 5θ = 90° – 45° = 45°
∴ θ = 9°Correct Option: B
tan (2θ + 45°) = cot 3θ
= tan (90° – 3θ )
⇒ 2θ + 45° = 90° – 3θ
⇒ 5θ = 90° – 45° = 45°
∴ θ = 9°
