Trigonometry
- If sin (A – B) = sin A cos B – cosA sinB, then sin 15° will be
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Let, A = 45°
B = 30°
sin (A – B)
= sin A. cos B – cos A. sin B
⇒ sin (45° – 30°)
= sin 45°. cos 30° – cos 45°. sin 30°
⇒ sin 15°= 1 × √3 - 1 × 1 √2 2 √2 2 = √3 - 1 = √3 - 1 2√2 2√2 2√2
Correct Option: D
Let, A = 45°
B = 30°
sin (A – B)
= sin A. cos B – cos A. sin B
⇒ sin (45° – 30°)
= sin 45°. cos 30° – cos 45°. sin 30°
⇒ sin 15°= 1 × √3 - 1 × 1 √2 2 √2 2 = √3 - 1 = √3 - 1 2√2 2√2 2√2
- If sec x + cos x = 2, then the value of sec16 x + cos16 x will be
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sec x + cos x = 2
⇒ 1 + cosx = 2 cos x ⇒ 1 + cos²x = 2 cos x
⇒ cos²x + 1 = 2 cosx
⇒ cos²x – 2 cosx + 1 = 0
⇒ (cosx – 1)² = 0
⇒ cos –1 = 0
⇒ cos = 1
∴ secx = 1
∴ sec16x + cos16x = 1 + 1 = 2Correct Option: B
sec x + cos x = 2
⇒ 1 + cosx = 2 cos x ⇒ 1 + cos²x = 2 cos x
⇒ cos²x + 1 = 2 cosx
⇒ cos²x – 2 cosx + 1 = 0
⇒ (cosx – 1)² = 0
⇒ cos –1 = 0
⇒ cos = 1
∴ secx = 1
∴ sec16x + cos16x = 1 + 1 = 2
- If sin4θ + cos4θ = 2 sin2θ cos2θ, θ is an acute angle, then the value of tan θ is
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sin4 θ + cos4 θ = 2 sin² θ. cos²θ
⇒ sin4 θ + cos4 θ
– 2 sin² θ. cos² θ = 0
⇒ (sin² θ – cos² θ)² = 0
⇒ sin² θ – cos² θ = 0
⇒ sin² θ = cos² θ
= tan² θ = 1
⇒ tan θ = + 1
∵ θ is acute angle.Correct Option: A
sin4 θ + cos4 θ = 2 sin² θ. cos²θ
⇒ sin4 θ + cos4 θ
– 2 sin² θ. cos² θ = 0
⇒ (sin² θ – cos² θ)² = 0
⇒ sin² θ – cos² θ = 0
⇒ sin² θ = cos² θ
= tan² θ = 1
⇒ tan θ = + 1
∵ θ is acute angle.
- The maximum value of sin2θ + cos2θ is
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Expression
= sin4θ + cos4θ
= (sin²θ)² + (cos²θ)²
= (sin²θ + cos²θ)² – 2 sin²θ.cos²θ.
= 1 – 2 sin²θ. cos²θ.= 1 - 4sin²θ . cos²θ 2
[∵ sin²θ = 2 sinθ . cosθ]= 1 - sin²2θ 2 = 1 - 1 - cos4θ 4
[∵ 1 – cos²θ = 2cos²θ]= 1 - 1 + cos4θ 4 4 = 1 - 1 + 1 = 1 4 4
(cos 4θ ≤1)
OR
The value of sin4 θ + cos4 θ will be
maximum if θ = 0°
∴ Required value = (sin0)4 + (cos0)4 = 0 + 1 = 1Correct Option: B
Expression
= sin4θ + cos4θ
= (sin²θ)² + (cos²θ)²
= (sin²θ + cos²θ)² – 2 sin²θ.cos²θ.
= 1 – 2 sin²θ. cos²θ.= 1 - 4sin²θ . cos²θ 2
[∵ sin²θ = 2 sinθ . cosθ]= 1 - sin²2θ 2 = 1 - 1 - cos4θ 4
[∵ 1 – cos²θ = 2cos²θ]= 1 - 1 + cos4θ 4 4 = 1 - 1 + 1 = 1 4 4
(cos 4θ ≤1)
OR
The value of sin4 θ + cos4 θ will be
maximum if θ = 0°
∴ Required value = (sin0)4 + (cos0)4 = 0 + 1 = 1
- Find the value of tan 4° tan 43° tan 47° tan 86°
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tan 86° = cot (90° – 86°) = cot 4°.
tan 47° = cot (90° – 47°) = cot 43°
7there4; (tan 4°. tan 86°) (tan 43°.tan 47°)
= (tan 4° : cot 4°) (tan 43°.cot 43°)
= 1 (∵ tanθ . cotθ = 1)Correct Option: B
tan 86° = cot (90° – 86°) = cot 4°.
tan 47° = cot (90° – 47°) = cot 43°
7there4; (tan 4°. tan 86°) (tan 43°.tan 47°)
= (tan 4° : cot 4°) (tan 43°.cot 43°)
= 1 (∵ tanθ . cotθ = 1)
