Trigonometry
- If sin θ + cos θ = √2 sin (90° – θ) then cot θ is equal to :
-
View Hint View Answer Discuss in Forum
sinθ + cosθ = √2 sin (90° – θ)
⇒ sinθ + cosq = √2 cosθ
⇒ √2 cosθ – cosθ = sinθ
⇒ cosθ (√2) - 1 = sinθ⇒ cos θ = 1 sin θ √2 - 1 ⇒ cot θ 1 × √2 + 1 √2 - 1 √2 + 1 = √2 + 1 = √2 + 1 2 - 1
Correct Option: D
sinθ + cosθ = √2 sin (90° – θ)
⇒ sinθ + cosq = √2 cosθ
⇒ √2 cosθ – cosθ = sinθ
⇒ cosθ (√2) - 1 = sinθ⇒ cos θ = 1 sin θ √2 - 1 ⇒ cot θ 1 × √2 + 1 √2 - 1 √2 + 1 = √2 + 1 = √2 + 1 2 - 1
- The value of the following is :
(tan 20°)² + (cot 20°)² + 2tan 15°. tan 45°. tan 75° (cosec 70°)² (sec 70°)²
-
View Hint View Answer Discuss in Forum
tan 20° = tan (90° – 70°) = cot 70°
∴ cot 20° = tan 70°
tan 15° = tan (90° – 75°) = cot 75°
∴ Expression = cot² 70°. sin²70° + tan²70°. cos²70° + 2 cot 75°. tan 75°. tan 45°= cos² 70° . sin² 70° + sin² 70° sin² 70° cos² 70°
cos² 70° + 2 × 1 × 1
= cos²70° + sin²70° + 2 = 1 + 2 = 3
[∵ sinθ . cosecθ = 1 ; cosθ. secθ = 1; tanθ . cotθ = 1]Correct Option: C
tan 20° = tan (90° – 70°) = cot 70°
∴ cot 20° = tan 70°
tan 15° = tan (90° – 75°) = cot 75°
∴ Expression = cot² 70°. sin²70° + tan²70°. cos²70° + 2 cot 75°. tan 75°. tan 45°= cos² 70° . sin² 70° + sin² 70° sin² 70° cos² 70°
cos² 70° + 2 × 1 × 1
= cos²70° + sin²70° + 2 = 1 + 2 = 3
[∵ sinθ . cosecθ = 1 ; cosθ. secθ = 1; tanθ . cotθ = 1]
- The value of the following is
sin 47° 
+ cos 43° - 4cos²45° cos 43° sin 47°
-
View Hint View Answer Discuss in Forum
sin47° = sin(90° – 43°) = cos43°
= 1 + 1 – 4 × 1 = 2 – 2 = 0 2
Correct Option: B
sin47° = sin(90° – 43°) = cos43°
= 1 + 1 – 4 × 1 = 2 – 2 = 0 2
- If 0° < θ < 90° and cosecθ = cot²θ, then the value of the expression cosec4θ – 2cosec4θ + cot²θ is equal to:
-
View Hint View Answer Discuss in Forum
cosecθ = cot²θ
⇒ cosecθ = cosec²θ – 1
⇒ cosec²θ – cosecθ = 1 .....(i)
Expression
= cosec4θ – 2cosec3θ + cot²θ
= cosec4θ – cosec3θ – cosec3θ + cosecθ
= cosec²θ (cosec²θ – cosecθ) – cosecθ (cosec²θ–1)
= cosec²θ – cosec²θ = 0Correct Option: B
cosecθ = cot²θ
⇒ cosecθ = cosec²θ – 1
⇒ cosec²θ – cosecθ = 1 .....(i)
Expression
= cosec4θ – 2cosec3θ + cot²θ
= cosec4θ – cosec3θ – cosec3θ + cosecθ
= cosec²θ (cosec²θ – cosecθ) – cosecθ (cosec²θ–1)
= cosec²θ – cosec²θ = 0
- If 4sin²θ – 1 = 0 and angle θ is less than 90°, the value of cos²θ + tan²θ is : (Take 0° < θ < 90°)
-
View Hint View Answer Discuss in Forum
4 sin²θ – 1 = 0
⇒ 4 sin²θ = 1⇒ sin²θ = 1 4 ⇒ sinθ = 1 (∵ θ < 90°) 2
∴sinθ = sin30°
⇒ θ = 30°
∴ cos²θ + tan²θ = cos²30° + tan²30°
= 9 + 4 = 13 12 12
Correct Option: B
4 sin²θ – 1 = 0
⇒ 4 sin²θ = 1⇒ sin²θ = 1 4 ⇒ sinθ = 1 (∵ θ < 90°) 2
∴sinθ = sin30°
⇒ θ = 30°
∴ cos²θ + tan²θ = cos²30° + tan²30°= 9 + 4 = 13 12 12
