Given tan A + cot A = 2. Since \(\cot A = \frac{1}{\tan A}\), we can rewrite the given equation as: \(\tan A + \frac{1}{\tan A} = 2\)
Let \(x = \tan A\).
Then \(x + \frac{1}{x} = 2\).
Multiplying by \(x\) gives \(x^2 + 1 = 2x\), or \(x^2 - 2x + 1 = 0\).
This factors as \((x-1)^2 = 0\), so \(x=1\). Thus, \(\tan A = 1\), which implies \(\cot A = \frac{1}{\tan A} = 1\).
Now, tan⁴ A + cot⁴ A = 1⁴ + 1⁴ = 1 + 1 = 2.
Answer: (A) 2
The direction cosines of two lines are connected by the relations \( 1 + m - n = 0 \) and \( lm - 2mn + nl = 0 \). If \( \theta \) is the acute angle between those lines, then \( \cos \theta = \) ?