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Answers
Answer:
Given Quadratic equation:
√3cot²θ-4cotθ+√3=0
Splitting the middle term, we get
=> √3cot²θ-3cotθ-1cotθ+√3=0
=> √3cotθ(cotθ-√3)-1(cotθ-√3)=0
=> (cotθ-√3)(√3cotθ-1)=0
=> cotθ-√3 = 0 Or √3cotθ-1=0
=> cotθ = √3 Or cotθ = 1/√3
case 1:
cotθ = √3 => tanθ = 1/√3
Now,
cot²θ + tan²θ
= (√3)²+(1/√3)²
= 3 + 1/3
= (9+1)/3
= 10/3
case 2:
If cotθ = 1/√3 => tanθ = √3
Now,
cot²θ+tan²θ
= (1/√3)²+(√3)²
= 1/3 + 3/1
= (1+9)/3
= 10/3
Therefore,
cot²θ + tan²θ = 10/3
Answer:
Answer:
Given Quadratic equation:
√3cot²θ-4cotθ+√3=0
Splitting the middle term, we get
=> √3cot²θ-3cotθ-1cotθ+√3=0
=> √3cotθ(cotθ-√3)-1(cotθ-√3)=0
=> (cotθ-√3)(√3cotθ-1)=0
=> cotθ-√3 = 0 Or √3cotθ-1=0
=> cotθ = √3 Or cotθ = 1/√3
case 1:
cotθ = √3 => tanθ = 1/√3
Now,
cot²θ + tan²θ
= (√3)²+(1/√3)²
= 3 + 1/3
= (9+1)/3
= 10/3
case 2:
If cotθ = 1/√3 => tanθ = √3
Now,
cot²θ+tan²θ
= (1/√3)²+(√3)²
= 1/3 + 3/1
= (1+9)/3
= 10/3
Therefore,
cot²θ + tan²θ = 10/3