If 3 tan X + cot X = 2√3, then find the value of 6

Solution

3 tan X + cot X = 2√3

We know that, cot A = (1/tan A)

3 tan X + (1/tan X) = 2√3

Or, 3 tan² X + 1 = 2√3 tan X

Or, 3 tan² X - 2√3 tan X + 1 = 0

Or, 3 tan² X - √3 tan X - √3tan X + 1 = 0

Or, √3 tan X X (√3 tan X - 1) + ( - 1) X (√3 tan X - 1) = 0

Or, (√3 tan X - 1) X (√3 tan X - 1) = 0

Or, (√3 tan X - 1) ² = 0

Or, (√3 tan X - 1) = 0

Or, tan X = (1/√3) = tan30^o [because, tan30^o = (1/√3) ]

So, X = 30^o
6 tan² X + 2 cot² X = 6 tan² 30^o + 2 cot² 30^o

= 6 X (1/√3) ² + 2 X (√3) ²

= 6 X (1/3) + 2 X 3 = 2 + 6 = 8