Factor the trigonometric expression completely. \[ 2 \cot ^{2} W+13 \cot W+21 \]

Step 1 ：Given expression is \(2\cot^2(W) + 13\cot(W) + 21\)

Step 2 ：We need to find two numbers such that their product is \(2*21 = 42\) and their sum is 13

Step 3 ：The numbers 6 and 7 satisfy these conditions because \(6*7 = 42\) and \(6+7 = 13\)

Step 4 ：So, we can write the middle term as the sum of \(6\cot(W)\) and \(7\cot(W)\)

Step 5 ：\(2\cot^2(W) + 13\cot(W) + 21 = 2\cot^2(W) + 6\cot(W) + 7\cot(W) + 21\)

Step 6 ：Now, we can factor by grouping: \(2\cot(W)[\cot(W) + 3] + 7[\cot(W) + 3]\)

Step 7 ：Now, take out the common factor \([\cot(W) + 3]\)

Step 8 ：So, the factorized form of the given expression is \((2\cot(W) + 7)(\cot(W) + 3)\)

Step 9 ：Final Answer: \(\boxed{(2\cot(W) + 7)(\cot(W) + 3)}\)