Question
If 3 tan X + cot X = 2√3, then find the value of 6
tan2 X + 2 cot2 X.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) = tan30o [because, tan30o = (1/√3) ]
So, X = 30o
6 tan2 X + 2 cot2 X = 6 tan2 30o + 2 cot2 30o
= 6 X (1/√3) 2 + 2 X (√3) 2
= 6 X (1/3) + 2 X 3 = 2 + 6 = 8
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