If \(4 \cot ^{2} \alpha-16 \cot \alpha+15<0\) and \(\alpha \in R\) then cota lies in interval (a) \([(3 / 2),(5 / 2)]\) (b) \([0,(3 / 2)]\) (c) \([0,(5 / 2)]\) (d) \([(5 / 2), \infty]\)

The given inequality is \(4\cot^2\alpha - 16\cot\alpha + 15 < 0\). Let's rewrite the inequality as it is easier to factorize a quadratic expression for positive values. \(4\cot^2\alpha - 16\cot\alpha + 15 < 0\) To factor, we can rewrite the equation as: \(4(\cot\alpha)^2 - 16(\cot\alpha) + 15 = 4(\cot\alpha - 5)(\cot\alpha - 3)\) Now the inequality can be written as: \(4(\cot\alpha - 5)(\cot\alpha - 3) < 0\)

Since we have factored the inequality, we can now find the critical points which make the inequality 0: \(\cot\alpha = 5, \cot\alpha = 3\) These critical points create intervals. Interval 1: \((-\infty, 3)\) Interval 2: \((3, 5)\) Interval 3: \((5, \infty)\)

Now we need to test the inequality for one point in each of these intervals and determine which interval(s) the inequality holds true. Interval 1: \(\cot\alpha = 0\) since 0 is in the interval (-∞, 3), we will check if the inequality is true for \(\alpha\) such that \(\cot\alpha = 0\) \(4(0 - 5)(0 - 3) = 60 > 0\) The inequality does not hold true for this interval. Interval 2: \(\cot\alpha = 4\) since 4 is in the interval (3, 5), we will check if the inequality is true for \(\alpha\) such that \(\cot\alpha = 4\) \(4(4 - 5)(4 - 3) = -4 < 0\) The inequality holds true for this interval. Interval 3: \(\cot\alpha = 6\) since 6 is in the interval (5, ∞), we will check if the inequality is true for \(\alpha\) such that \(\cot\alpha = 6\) \(4(6 - 5)(6 - 3) = 12 > 0\) The inequality does not hold true for this interval. The solution is in the interval (3, 5) or \(\cot\alpha \in [(5/2)^{-1},(3/2)^{-1}]\). The correct answer is option (a).