Trigonometry
- If 0° < q < 90°, the value of sin q + cos q is
-
View Hint View Answer Discuss in Forum
Z = sinθ + cosθ
⇒ Z² = sin²θ + cos²θ + 2 sinθ.cosθ
= 1 + 2 sinθ. cosθ
∵ 0 < θ < 90°
∴ sinθ < 1; cosθ < 1
∴ 2sinθ . cosθ < 1
and 2sinθcosθ > 1
⇒ Z² < 2
⇒ Z < √2Correct Option: B
Z = sinθ + cosθ
⇒ Z² = sin²θ + cos²θ + 2 sinθ.cosθ
= 1 + 2 sinθ. cosθ
∵ 0 < θ < 90°
∴ sinθ < 1; cosθ < 1
∴ 2sinθ . cosθ < 1
and 2sinθcosθ > 1
⇒ Z² < 2
⇒ Z < √2
- If A = tan 11° tan 29°, B = 2 cot 61° cot 79°, then :
-
View Hint View Answer Discuss in Forum
A = tan11° . tan 29°
B = 2 cot 61° . cot 79°
= 2 cot (90° – 29°) cot (90° – 11°)
= 2 tan 29° . tan 11°
[∵ cot (90° – θ) = tan θ]
= 2 ACorrect Option: C
A = tan11° . tan 29°
B = 2 cot 61° . cot 79°
= 2 cot (90° – 29°) cot (90° – 11°)
= 2 tan 29° . tan 11°
[∵ cot (90° – θ) = tan θ]
= 2 A
- The value of (sin 39° / cos 51°) + 2 tan 11° tan 31° tan 45° tan 59° tan 79° – 3 (sin² 21° + sin² 69°) is :
-
View Hint View Answer Discuss in Forum
= sin 39° + 2 tan11°. tan (90° – 11°).tan31°. tan (90° – 59°). 1 – 3 (sin² 21° + sin² (90° – 21°) cos(90° - 39°) = sin 39° + 2 tan11°. cot11°. tan31°.cot31°–3 (sin² 21°+cos² 21°) = 1 + 2 – 3 = 0 sin 39°
[tan θ .cot θ = 1 , sin² θ + cos²θ = 1 ]Correct Option: D
= sin 39° + 2 tan11°. tan (90° – 11°).tan31°. tan (90° – 59°). 1 – 3 (sin² 21° + sin² (90° – 21°) cos(90° - 39°) = sin 39° + 2 tan11°. cot11°. tan31°.cot31°–3 (sin² 21°+cos² 21°) = 1 + 2 – 3 = 0 sin 39°
[tan θ .cot θ = 1 , sin² θ + cos²θ = 1 ]
- sin²5° + sin²10° + sin²15° + .... + sin²85° + sin²90° is equal to
-
View Hint View Answer Discuss in Forum
sin θ = cos (90° – θ);
sin (90° – θ) = cos θ
∴ sin 85° = sin (90° – 5°) = cos 5°
∴ (sin² 5° + sin² 85°) + (sin² 10° + sin² 80°) + ..... to 8 terms + sin² 45° + sin² 90°= 8 × 1 + 1 + 1 = 9 1 2 2
Correct Option: D
sin θ = cos (90° – θ);
sin (90° – θ) = cos θ
∴ sin 85° = sin (90° – 5°) = cos 5°
∴ (sin² 5° + sin² 85°) + (sin² 10° + sin² 80°) + ..... to 8 terms + sin² 45° + sin² 90°= 8 × 1 + 1 + 1 = 9 1 2 2
- sin² 5° + sin² 6° + ... + sin² 84° + sin² 85° = ?
-
View Hint View Answer Discuss in Forum
Let the number of terms be n, then
By t 2 = a + (n – 1)d
85 = 5 + (n –1)
⇒ n – 1 = 85 – 5 = 80
⇒ n = 81
∴ sin²5° + sin²6°+ ... + sin²45° + ... + sin² 84° + sin² 85°
= (sin²5° + sin²85°) + (sin²6° + ... + sin² 84°) + ..... + to (40 terms) + sin²45°
= (sin²5° + cos25°) + (sin²6° + ... + cos² 6°) + ..... + to 40 terms + sin²45°
Correct Option: B
Let the number of terms be n, then
By t 2 = a + (n – 1)d
85 = 5 + (n –1)
⇒ n – 1 = 85 – 5 = 80
⇒ n = 81
∴ sin²5° + sin²6°+ ... + sin²45° + ... + sin² 84° + sin² 85°
= (sin²5° + sin²85°) + (sin²6° + ... + sin² 84°) + ..... + to (40 terms) + sin²45°
= (sin²5° + cos25°) + (sin²6° + ... + cos² 6°) + ..... + to 40 terms + sin²45°
