Trigonometry
- If cos2x + cos4x = 1, then tan2x + tan4x = ?
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cos2x + cos4x = 1
⇒ cos4x = 1 – cos2x = sin2x∴ tan2x + tan4x = sin2x + sin4x cos2x cos4x ⇒ tan2x + tan4x = cos4x + sin4x cos2x sin2x
Hence , tan2x + tan4x = cos2x + sin2x = 1
Correct Option: B
cos2x + cos4x = 1
⇒ cos4x = 1 – cos2x = sin2x∴ tan2x + tan4x = sin2x + sin4x cos2x cos4x ⇒ tan2x + tan4x = cos4x + sin4x cos2x sin2x
Hence , tan2x + tan4x = cos2x + sin2x = 1
- If cos 27° = x, the value of tan63° is
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cos 27° = x
⇒ cos (90° – 63°) = x
⇒ sin 63° = x
∴ cos 63° = √1 - sin² 63° = √1 - x²∴ tan 63 = sin 63° = x cos 63° √1 - x² Correct Option: A
cos 27° = x
⇒ cos (90° – 63°) = x
⇒ sin 63° = x
∴ cos 63° = √1 - sin² 63° = √1 - x²∴ tan 63 = sin 63° = x cos 63° √1 - x²
- If tan (5x – 10°) = cot (5y + 20°), then the value of (x + y) is
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tan (90° – θ) = cotθ
∴ tan (5x – 10°) = cot (5y + 20°)
⇒ tan (5x – 10°) = tan {90° – (5y + 20°)}
⇒ 5x – 10° = 90° – (5y + 20°)
⇒ 5x – 10° = 90° – 5y – 20°
⇒ 5x + 5y = 70° + 10°
⇒ 5 (x + y) = 80°⇒ x + y = 80° = 16° 5
Correct Option: B
tan (90° – θ) = cotθ
∴ tan (5x – 10°) = cot (5y + 20°)
⇒ tan (5x – 10°) = tan {90° – (5y + 20°)}
⇒ 5x – 10° = 90° – (5y + 20°)
⇒ 5x – 10° = 90° – 5y – 20°
⇒ 5x + 5y = 70° + 10°
⇒ 5 (x + y) = 80°⇒ x + y = 80° = 16° 5
- If cos 20° = m and cos 70° = n, then the value of m2 + n2 is
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cos 20° = m and cos 70° = n
∴ m2 + n2 = cos2 20°+ cos2 70°
= cos2 (90° – 70°) + cos2 70°
⇒ sin2 70° + cos2 70° = 1Correct Option: A
cos 20° = m and cos 70° = n
∴ m2 + n2 = cos2 20°+ cos2 70°
= cos2 (90° – 70°) + cos2 70°
⇒ sin2 70° + cos2 70° = 1
- The value of cos 1° cos 2°cos 3° ... cos 180° is
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∵ cos 90° = 0
∴ cos1°.cos2°.... cos 180° = 0Correct Option: A
∵ cos 90° = 0
∴ cos1°.cos2°.... cos 180° = 0
