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If 1 + cos²θ = 3 sinθ cosθ, then the integral value of
cot θ 
0 < θ < π 
is 2
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- 1
- 2
- 0
- 3
Correct Option: A
1 + cos²θ = 3 sinθ . cosθ
Dividing both sides by sin²θ
| + | = | 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 = | or 1 | 2 |
