Question

Please solve this and give the correct answer tomorrow is my examination......​

Answer 1

$\bf\large\underline\pink{Answer:-}$

Solution:

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

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