Trigonometry
- If x cosθ – sinθ = 1, then x² + (1 +x² ) sinq equals
-
View Hint View Answer Discuss in Forum
x cosθ – sinθ = 1
⇒ x cosθ = 1 + sinθ⇒ x = 1 + sin θ cos θ cos θ
⇒ x = secθ + tanθ --- (i)
∵ sec²θ – tan²θ = 1
⇒ (secθ + tanθ) (secθ – tanθ) =1⇒ secθ – tanθ = 1 (ii) x
From equation (i) + (ii),2secθ = x + 1 = x² + 1 x x ⇒ secθ = x² + 1 2x
From equation (i) – (ii),2tanθ = x – 1 = x² - 1 x x ∴ tanθ = x² - 1 2x ∴ sinθ = tanθ secθ = x² - 1 × 2x = x² - 1 2x x² + 1 x² + 1
∴ Expression = x² – (1 + x² ) sinθ= x² - (1 + x²) × x² - 1 = x² - x² + 1 = 1 x² + 1
Note : In the original equation x² + (1 + x² ) sinθ has been given that seems incorrect
Correct Option: B
x cosθ – sinθ = 1
⇒ x cosθ = 1 + sinθ⇒ x = 1 + sin θ cos θ cos θ
⇒ x = secθ + tanθ --- (i)
∵ sec²θ – tan²θ = 1
⇒ (secθ + tanθ) (secθ – tanθ) =1⇒ secθ – tanθ = 1 (ii) x
From equation (i) + (ii),2secθ = x + 1 = x² + 1 x x ⇒ secθ = x² + 1 2x
From equation (i) – (ii),2tanθ = x – 1 = x² - 1 x x ∴ tanθ = x² - 1 2x ∴ sinθ = tanθ secθ = x² - 1 × 2x = x² - 1 2x x² + 1 x² + 1
∴ Expression = x² – (1 + x² ) sinθ= x² - (1 + x²) × x² - 1 = x² - x² + 1 = 1 x² + 1
Note : In the original equation x² + (1 + x² ) sinθ has been given that seems incorrect
- If sin θ + cos θ = √2 cos θ, then the value of cot θ is
-
View Hint View Answer Discuss in Forum
sinθ + cosθ =√2 cos θ
⇒ √2 cos θ - cos θ= sinθ
⇒ cosθ (√2 - 1) = sinθ⇒ cosθ = 1 sinθ √2 - 1 = √2 + 1 = √2 + 1 (√2 - 1)(√2 + 1)
cotθ = √2 + 1
Correct Option: A
sinθ + cosθ =√2 cos θ
⇒ √2 cos θ - cos θ= sinθ
⇒ cosθ (√2 - 1) = sinθ⇒ cosθ = 1 sinθ √2 - 1 = √2 + 1 = √2 + 1 (√2 - 1)(√2 + 1)
cotθ = √2 + 1
- If cos4θ – sin4θ = (2 / 3) , then the value of 1 – 2 sin2 θ is
-
View Hint View Answer Discuss in Forum
cos4θ – sin4θ = 2 3 ⇒ (cos²θ + sin²θ) (cos²θ– sin²θ) = 2 3
[ ∵ cos²θ + sin²θ = 1]⇒ cos²θ + sin²θ = 2 3 ⇒ 1 - sin²θ - sin²θ = 2 3 ⇒ 1 - 2 sin²θ = 2 3
Correct Option: A
cos4θ – sin4θ = 2 3 ⇒ (cos²θ + sin²θ) (cos²θ– sin²θ) = 2 3
[ ∵ cos²θ + sin²θ = 1]⇒ cos²θ + sin²θ = 2 3 ⇒ 1 - sin²θ - sin²θ = 2 3 ⇒ 1 - 2 sin²θ = 2 3
-
The value of cot 30° - cot 75° is equal to tan 15° - tan 60°
-
View Hint View Answer Discuss in Forum
cot 30° - cot 75° tan 15° - tan 60° = cot (90° - 60°) - cot (90° - 15°) tan 15° - tan 60° = cot 60° - tan 15° = - 1 tan 15° - tan 60°
Correct Option: A
cot 30° - cot 75° tan 15° - tan 60° = cot (90° - 60°) - cot (90° - 15°) tan 15° - tan 60° = cot 60° - tan 15° = - 1 tan 15° - tan 60°
- If sin θ + cos θ = p and sec θ + cosec θ = θ, then the value of θ (p² – 1) is
-
View Hint View Answer Discuss in Forum
sin θ + cos θ = p
sec θ + cosec θ = q⇒ 1 + 1 = q cos θ sinθ ⇒ sinθ + cosθ = q sinθ . cosθ ∴ q(p² - 1) = 
sin θ + cos θ 
((sinθ + cosθ)² - 1) sinθ . cosθ = sinθ + cosθ . (sin²θ + cos²θ + 2sinθ.cosθ - 1) sinθ . cosθ = sinθ + cosθ . 2sinθ . cosθ sinθ . cosθ
= 2pCorrect Option: C
sin θ + cos θ = p
sec θ + cosec θ = q⇒ 1 + 1 = q cos θ sinθ ⇒ sinθ + cosθ = q sinθ . cosθ ∴ q(p² - 1) = sin θ + cos θ ((sinθ + cosθ)² - 1) sinθ . cosθ = sinθ + cosθ . (sin²θ + cos²θ + 2sinθ.cosθ - 1) sinθ . cosθ = sinθ + cosθ . 2sinθ . cosθ sinθ . cosθ
= 2p
