Trigonometry
- In ∆ABC, ∠C = 90° and AB = c, BC = a, CA = b; then the value of (cosec B – cos A) is
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In ∆ABC, AB² = AC² + BC²
⇒ c² = a² + b² ........ (i)
From ∆ABC,cosec B = AB = c ....... (ii) AC b cos A = AC = b AB c ∴ cosec B – cos A = c = b b c = c² - b² = a² bc bc
Correct Option: C
In ∆ABC, AB² = AC² + BC²
⇒ c² = a² + b² ........ (i)
From ∆ABC,cosec B = AB = c ....... (ii) AC b cos A = AC = b AB c ∴ cosec B – cos A = c = b b c = c² - b² = a² bc bc
- If tan θ – cot θ = 0 and θ is positive acute angle, then the value of
tan(θ + 15°) is tan(θ - 15°)
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tanθ – cotθ = 0
⇒ tanθ = cotθ
⇒ tanθ = tan (90° – θ)
⇒ θ = 90° – θ
⇒ 2θ = 90°
⇒ θ = 45°∴ tan(θ + 15°) tan(θ - 15°) = tan(45° + 15°) = tan 60° tan(45° - 15°) tan 30° 
= √3 = √3 × √3 = 3 (1 / √3)
Correct Option: A
tanθ – cotθ = 0
⇒ tanθ = cotθ
⇒ tanθ = tan (90° – θ)
⇒ θ = 90° – θ
⇒ 2θ = 90°
⇒ θ = 45°∴ tan(θ + 15°) tan(θ - 15°) = tan(45° + 15°) = tan 60° tan(45° - 15°) tan 30° = √3 = √3 × √3 = 3 (1 / √3)
- The value of cot 41°. cot 42° . cot 43°. cot 44°. cot 45° . cot 46° . cot 47° . cot 48° . cot 49°
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Expression = (cot 41° . cot 49°). (cot 42° . cot 48°) (cot 43° . cot 47°) . (cot 44° . cot 46°) . cot 45°
= cot 41° . tan (90° – 49°) . cot 42° . tan (90° – 48°) . cot 43° . tan (90° – 47°) . cot 44° . tan (90° – 46°).1
= (cot 41° . tan 41°) (cot 42° . tan 42°) . (cot 43° . tan 43°). (cot 44°
. tan 44°) . 1 = 1
[∵ tan (90° – θ) = cotθ; tanθ. cotθ = 1]Correct Option: A
Expression = (cot 41° . cot 49°). (cot 42° . cot 48°) (cot 43° . cot 47°) . (cot 44° . cot 46°) . cot 45°
= cot 41° . tan (90° – 49°) . cot 42° . tan (90° – 48°) . cot 43° . tan (90° – 47°) . cot 44° . tan (90° – 46°).1
= (cot 41° . tan 41°) (cot 42° . tan 42°) . (cot 43° . tan 43°). (cot 44°
. tan 44°) . 1 = 1
[∵ tan (90° – θ) = cotθ; tanθ. cotθ = 1]
- If x = a sin θ – b cos θ, y = a cos θ + b sin θ, then which of the following is true?
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x = a sinθ – b cosθ --- (i)
y = a cosθ + b sinθ - (ii)
On squaring and adding both the equations,
x² + y² = (a sinθ – b cosθ)² + (a cosθ + b sinθ)²
= a² sin²θ + b² cos²θ – 2ab sinθ . cosθ + a² cos²θ + b² sin²θ + 2ab sinθ . cosθ
= a² (sin²θ + cos²θ) + b² (cos²θ + sin²θ)
= a² + b² [∵ sin²θ + cos²θ = 1]Correct Option: D
x = a sinθ – b cosθ --- (i)
y = a cosθ + b sinθ - (ii)
On squaring and adding both the equations,
x² + y² = (a sinθ – b cosθ)² + (a cosθ + b sinθ)²
= a² sin²θ + b² cos²θ – 2ab sinθ . cosθ + a² cos²θ + b² sin²θ + 2ab sinθ . cosθ
= a² (sin²θ + cos²θ) + b² (cos²θ + sin²θ)
= a² + b² [∵ sin²θ + cos²θ = 1]
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If sec θ – tan θ = 1 , the value of sec θ . tan θ is √3
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secθ – tanθ = 1 ....(i) √3
∵ sec²θ – tan²θ = 1
⇒ (secθ + tanθ) (secθ – tanθ) = 1
⇒ secθ + tanθ = 3 ....(ii)
On adding equations (i) and (ii)2secθ = √3 + 1 √3 = 3 + 1 = 4 √3 √3 ⇒ secθ = 2 √3
Again, by equation (ii) – (i),2 tanθ = √3 - 2 √3 = 3 - 1 = 2 √3 √3 ⇒ tanθ = 2 √3
∴ secθ . tanθ= 2 × 1 = 2 √3 √3 3
Correct Option: A
secθ – tanθ = 1 ....(i) √3
∵ sec²θ – tan²θ = 1
⇒ (secθ + tanθ) (secθ – tanθ) = 1
⇒ secθ + tanθ = 3 ....(ii)
On adding equations (i) and (ii)2secθ = √3 + 1 √3 = 3 + 1 = 4 √3 √3 ⇒ secθ = 2 √3
Again, by equation (ii) – (i),2 tanθ = √3 - 2 √3 = 3 - 1 = 2 √3 √3 ⇒ tanθ = 2 √3
∴ secθ . tanθ= 2 × 1 = 2 √3 √3 3
