Trigonometry
- Find the value of 1 – 2 sin²θ + sin4θ.
-
View Hint View Answer Discuss in Forum
1 – 2 sin²θ + sin4 θ
= (1 – sin²θ)² = (cos²θ)² = cos4θCorrect Option: B
1 – 2 sin²θ + sin4 θ
= (1 – sin²θ)² = (cos²θ)² = cos4θ
- The simplest value of cot 9° cot 27° cot 63° cot 81° is
-
View Hint View Answer Discuss in Forum
Expression
= cot 9°. cot 27°. cot 63° . cot 81°
= cot 9°. cot 27°. cot (90° – 27°) . cot (90° – 9°)
= cot 9° . cot 27° . tan 27° . tan 9° [tan (90° – θ)
= cot θ; cot (90° – θ) = tan θ ]
= cot 9° . tan 9° . cot 27° tan 27°
= 1 [tan θ . cot θ = 1]Correct Option: B
Expression
= cot 9°. cot 27°. cot 63° . cot 81°
= cot 9°. cot 27°. cot (90° – 27°) . cot (90° – 9°)
= cot 9° . cot 27° . tan 27° . tan 9° [tan (90° – θ)
= cot θ; cot (90° – θ) = tan θ ]
= cot 9° . tan 9° . cot 27° tan 27°
= 1 [tan θ . cot θ = 1]
- If (1 + sin A) (1 + sin B) (1 + sin C) = (1 – sin A) (1 – sin B) (1 – sin C), 0 < A, B, C < (π / 2) then each side is equal to
-
View Hint View Answer Discuss in Forum
(1 + sin A) (1 + sin B) (1 + sin C) = (1 – sin A) . (1 – sin B) (1 – sin C) = x (Let)
∴ x . x = (1 + sin A) (1 + sin B) (1 + sin C) (1 – sin A) (1 – sin B) (1 – sin C)
⇒ x² = (1 – sin² A) (1 – sin² B) (1 – sin² C)
⇒ x² = cos²A . cos²B . cos²C
⇒x = ± cos A . cos B . cos C
∵ 0 < A, B, C < π 2
∴ x = cos A . cos B . cos CCorrect Option: B
(1 + sin A) (1 + sin B) (1 + sin C) = (1 – sin A) . (1 – sin B) (1 – sin C) = x (Let)
∴ x . x = (1 + sin A) (1 + sin B) (1 + sin C) (1 – sin A) (1 – sin B) (1 – sin C)
⇒ x² = (1 – sin² A) (1 – sin² B) (1 – sin² C)
⇒ x² = cos²A . cos²B . cos²C
⇒x = ± cos A . cos B . cos C
∵ 0 < A, B, C < π 2
∴ x = cos A . cos B . cos C
- The value of q, which satisfies the equation tan²θ + 3 = 3 secθ, 0° ≤ θ < 90° is
-
View Hint View Answer Discuss in Forum
tan²θ + 3 = 3 sec θ
⇒ sec²θ – 1 + 3 = 3 sec θ
⇒ sec²θ – 3 sec θ + 2 = 0
⇒ sec²θ – 2 sec θ – sec θ + 2 = 0
⇒ secθ (sec θ – 2) – 1 (sec θ – 2) = 0
⇒ (sec θ – 2) (sec θ – 1) = 0
⇒ secθ = 2 or 1
⇒ θ = 60° or 0°.Correct Option: D
tan²θ + 3 = 3 sec θ
⇒ sec²θ – 1 + 3 = 3 sec θ
⇒ sec²θ – 3 sec θ + 2 = 0
⇒ sec²θ – 2 sec θ – sec θ + 2 = 0
⇒ secθ (sec θ – 2) – 1 (sec θ – 2) = 0
⇒ (sec θ – 2) (sec θ – 1) = 0
⇒ secθ = 2 or 1
⇒ θ = 60° or 0°.
- If sin θ = 0.7, then cos q, 0 ≤ θ < 90°, is
-
View Hint View Answer Discuss in Forum
sin θ = 0.7
∴ cos θ
= √1 - sin² θ= √1 - (0.7)²
= √1 - 0.49= √0.51Correct Option: C
sin θ = 0.7
∴ cos θ
= √1 - sin² θ= √1 - (0.7)²
= √1 - 0.49= √0.51