Trigonometry
- Evaluate : 3 cos 80° cosec 10° + 2 cos 59° cosec 31°
-
View Hint View Answer Discuss in Forum
3 cos 80°. cosec 10° + 2 cos 59° . cosec 31°
= 3cos(90°–10°). cosec 10° + 2 cos(90°– 31°) . cosec 31°
= 3 sin 10°. cosec 10° + 2 sin 31°. cosec 31°
= 3 + 2 = 5
[ ∵ cos (90°– θ) = sinθ ; sinθ . cosecθ = 1]Correct Option: D
3 cos 80°. cosec 10° + 2 cos 59° . cosec 31°
= 3cos(90°–10°). cosec 10° + 2 cos(90°– 31°) . cosec 31°
= 3 sin 10°. cosec 10° + 2 sin 31°. cosec 31°
= 3 + 2 = 5
[ ∵ cos (90°– θ) = sinθ ; sinθ . cosecθ = 1]
- The value of 152 (sin 30° + 2 cos² 45° + 3 sin 30° + 4 cos² 45° + .... + 17 sin 30°+ 18 cos² 45°) is
-
View Hint View Answer Discuss in Forum
152 (sin 30° + 2 cos² 45° + 3 sin 30° + 4 cos² 45° + .... + 17 sin 30° + 18 cos² 45°)
It is an A.P. whose a = 1 , d = 1 , n = 18 2 2 
= 152 × 9 × 19 = 12996 and √12996 = 114 2
i.e. a perfect square of an integer.Correct Option: C
152 (sin 30° + 2 cos² 45° + 3 sin 30° + 4 cos² 45° + .... + 17 sin 30° + 18 cos² 45°)
It is an A.P. whose a = 1 , d = 1 , n = 18 2 2 = 152 × 9 × 19 = 12996 and √12996 = 114 2
i.e. a perfect square of an integer.
- Maximum value of (2 sin θ + 3 cos θ) is
-
View Hint View Answer Discuss in Forum
Maximum value of a sin θ + b
cos θ = √a² + b²
∴ Maximum value of 2 sin θ + 3
cos θ = √2² + 3² = √3Correct Option: B
Maximum value of a sin θ + b
cos θ = √a² + b²
∴ Maximum value of 2 sin θ + 3
cos θ = √2² + 3² = √3
- If sin (A – B) = (1 / 2) and cos (A + B) = (1 / 2) where A > B > 0 and A + B is an acute angle, then the value B is
-
View Hint View Answer Discuss in Forum
sin (A – B) = 1 = sin 30° 2
⇒ A - B = 30°Again, cos (A + B) = 1 = cos 30° 2
⇒ A + B = 60°
∴ A + B + A – B = 30° + 60° = 90°
⇒ 2 A = 90°
⇒ A = 45°
∴ A - B = 30°
⇒ B = A - 30° = 45° - 30° = 15°= 15 × π = π radian 180 12
Correct Option: B
sin (A – B) = 1 = sin 30° 2
⇒ A - B = 30°Again, cos (A + B) = 1 = cos 30° 2
⇒ A + B = 60°
∴ A + B + A – B = 30° + 60° = 90°
⇒ 2 A = 90°
⇒ A = 45°
∴ A - B = 30°
⇒ B = A - 30° = 45° - 30° = 15°= 15 × π = π radian 180 12
- If a sin θ + b cos θ = c then the value of a cos θ – b sin θ is :
-
View Hint View Answer Discuss in Forum
a sin θ + bcos θ = c ...(i)
a cos θ – b sin θ = x ...(ii)
Squaring both the equations and adding,
a² sin² θ + b² cos² θ + 2 ab
sin θ. cos θ + a² cos² θ + b² sin² θ – 2ab sin θ . cos θ = c² + x²
⇒ a² sin² θ + a² cos² θ + b² cos² θ + b² sin² θ = c² + x²
⇒ a² (sin² θ + cos² θ) + b² (cos² θ + sin² θ) = c² + x²
⇒ a² + b² = c² + x²
⇒ x² = a² + b² – c²
⇒ x = ± √a² + b² - c²Correct Option: B
a sin θ + bcos θ = c ...(i)
a cos θ – b sin θ = x ...(ii)
Squaring both the equations and adding,
a² sin² θ + b² cos² θ + 2 ab
sin θ. cos θ + a² cos² θ + b² sin² θ – 2ab sin θ . cos θ = c² + x²
⇒ a² sin² θ + a² cos² θ + b² cos² θ + b² sin² θ = c² + x²
⇒ a² (sin² θ + cos² θ) + b² (cos² θ + sin² θ) = c² + x²
⇒ a² + b² = c² + x²
⇒ x² = a² + b² – c²
⇒ x = ± √a² + b² - c²
