Trigonometry
- If secθ + tanθ = p, (p ≠ 0) then secθ is equal to
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secθ + tanθ = p .....(i)
∵ sec²θ – tan²θ = 1
⇒ (secθ + tanθ) (secθ – tanθ) = 1⇒ secθ – tanθ = 1 ....(ii) p
On adding both the equations2 secθ = p + 1 p ⇒ secθ = 1 
p + 1 
2 p
Correct Option: D
secθ + tanθ = p .....(i)
∵ sec²θ – tan²θ = 1
⇒ (secθ + tanθ) (secθ – tanθ) = 1⇒ secθ – tanθ = 1 ....(ii) p
On adding both the equations2 secθ = p + 1 p ⇒ secθ = 1 p + 1 2 p
- If 1 + cos²θ = 3 sinθ cosθ, then the integral value of
cot θ 0 < θ < π is 2
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1 + cos²θ = 3 sinθ . cosθ
Dividing both sides by sin²θ1 + cos²θ = 3sinθcosθ sin²θ sin²θ sin²θ
⇒ cosec²θ + cot²θ = 3 cot θ
⇒ 1 + cot²θ + cot²θ = 3 cotθ
⇒ 2 cot²θ – 3 cotθ + 1 = 0
⇒ 2 cot²θ – 2 cotθ – cotθ + 1 = 0
⇒ 2 cot²θ (cot θ – 1) – 1 (cot θ – 1) = 0
⇒ (2cot θ –1) (cot θ –1) = 0⇒ cot θ = 1 or 1 2
Correct Option: A
1 + cos²θ = 3 sinθ . cosθ
Dividing both sides by sin²θ1 + cos²θ = 3sinθcosθ sin²θ sin²θ sin²θ
⇒ cosec²θ + cot²θ = 3 cot θ
⇒ 1 + cot²θ + cot²θ = 3 cotθ
⇒ 2 cot²θ – 3 cotθ + 1 = 0
⇒ 2 cot²θ – 2 cotθ – cotθ + 1 = 0
⇒ 2 cot²θ (cot θ – 1) – 1 (cot θ – 1) = 0
⇒ (2cot θ –1) (cot θ –1) = 0⇒ cot θ = 1 or 1 2
- The value of the following is : 3(sin4θ + cos4θ) + 2(sin6θ + cos6θ) + 12sin2θ cos2θ
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Expression = 3(sin4 θ + cos4 θ) + 2(sin6 θ + cos6 θ) + 12 sin² θ. cos² θ
= 3{(sin² θ + cos² θ)² – 2 sin² θ . cos² θ} + 2{(sin² θ + cos² θ)3 – 3 sin² θ . cos² θ (sin² θ + cos² θ) + 12 sin² θ . cos² θ [∵ a² + b²
= (a + b)² – 2ab ; a3 + b3
= (a + b)3 – 3ab (a + b)]
= 3 (1 – 2 sin²θ. cos²θ) + 2 (1 – 3 sin²θ . cos²θ) + 12 sin²θ. cos²θ = 3 – 6 sin²θ . cos²θ + 2 – 6
sin²θ cos²θ + 12 sin²θ . cos²θ=5Correct Option: D
Expression = 3(sin4 θ + cos4 θ) + 2(sin6 θ + cos6 θ) + 12 sin² θ. cos² θ
= 3{(sin² θ + cos² θ)² – 2 sin² θ . cos² θ} + 2{(sin² θ + cos² θ)3 – 3 sin² θ . cos² θ (sin² θ + cos² θ) + 12 sin² θ . cos² θ [∵ a² + b²
= (a + b)² – 2ab ; a3 + b3
= (a + b)3 – 3ab (a + b)]
= 3 (1 – 2 sin²θ. cos²θ) + 2 (1 – 3 sin²θ . cos²θ) + 12 sin²θ. cos²θ = 3 – 6 sin²θ . cos²θ + 2 – 6
sin²θ cos²θ + 12 sin²θ . cos²θ=5
- If secθ + tanθ = 2 + √5 , then the value of sinθ is (0° ≤ θ ≤ 90°)
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sec θ + tan θ = 2 + 5
∵ sec²θ – tan²θ = 1
⇒ (sec θ + tan θ) (sec θ – tan θ) = 1⇒ sec θ – tan θ = 1 √5 + 2 = 1 × √5 - 2 = √5 - 2 √5 + 2 √5 - 2 5 - 4
= √5 - 2
∴ sec θ + tan θ + sec θ – tan θ = 2 + √5 + √5 - 2
⇒ 2 secθ = 2√5
⇒ secθ = √5 ....(i)
Again,
sec θ + tan θ – (sec θ – tan θ)
= 2 + √5 - √5 + 2
⇒ 2 tanθ = 4
⇒ tanθ = 2 ....(ii)∴ sin θ tan θ = 2 sec θ √5
Correct Option: B
sec θ + tan θ = 2 + 5
∵ sec²θ – tan²θ = 1
⇒ (sec θ + tan θ) (sec θ – tan θ) = 1⇒ sec θ – tan θ = 1 √5 + 2 = 1 × √5 - 2 = √5 - 2 √5 + 2 √5 - 2 5 - 4
= √5 - 2
∴ sec θ + tan θ + sec θ – tan θ = 2 + √5 + √5 - 2
⇒ 2 secθ = 2√5
⇒ secθ = √5 ....(i)
Again,
sec θ + tan θ – (sec θ – tan θ)
= 2 + √5 - √5 + 2
⇒ 2 tanθ = 4
⇒ tanθ = 2 ....(ii)∴ sin θ tan θ = 2 sec θ √5
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If secθ + tanθ = 2 51 then the value of sin θ is secθ - tanθ 79
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sec θ + tan θ = 2 51 sec θ - tan θ 79 158 + 51 = 209 79 79
By componendo and dividendo,sec θ + tan θ + sec θ - tan θ = 209 + 79 sec θ + tan θ - sec θ + tan θ 209 - 79 2 sec θ = 288 2 tan θ 130 sec θ = 144 tan θ 65 ∴ sinθ = tan θ = 65 sec θ 144
Correct Option: B
sec θ + tan θ = 2 51 sec θ - tan θ 79 158 + 51 = 209 79 79
By componendo and dividendo,sec θ + tan θ + sec θ - tan θ = 209 + 79 sec θ + tan θ - sec θ + tan θ 209 - 79 2 sec θ = 288 2 tan θ 130 sec θ = 144 tan θ 65 ∴ sinθ = tan θ = 65 sec θ 144
