3 cot^2A-4cotA+3=0, then find the value of cot^2A+tan^2A
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Given √3 cot2θ - 4cotθ + √3 = 0.
⇒ √3 Cot2θ - 3 Cot θ - Cot θ+ √3 = 0
⇒ √3 Cot θ( Cot θ - √3) - 1( Cot θ - √3) = 0
⇒ ( Cot θ - √3)(√3 Cot θ - 1) = 0
Cot θ = √3 and Cot θ = 1 / √3 .
let Cot θ = √3 and Tan θ = 1 / √3 .
Now Cot2θ +Tan2θ = 3 + 1 / 3 = 10 / 3.
Given √3 cot2θ - 4cotθ + √3 = 0.
⇒ √3 Cot2θ - 3 Cot θ - Cot θ+ √3 = 0
⇒ √3 Cot θ( Cot θ - √3) - 1( Cot θ - √3) = 0
⇒ ( Cot θ - √3)(√3 Cot θ - 1) = 0
Cot θ = √3 and Cot θ = 1 / √3 .
let Cot θ = √3 and Tan θ = 1 / √3 .
Now Cot2θ +Tan2θ = 3 + 1 / 3 = 10 / 3.
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