Question

√3cot^2o - 4coto+√3=0, then cot2o+ tan2o​

Answer 1

Step-by-step explanation:

√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.

Answer 2

Answer:

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