Trigonometry & Inverse Trigonometry
Modulus Inequalities
Grade 11
Question:
<p>Is \(|\tan x + \cot x| < |\tan x| + |\cot x|\) true for any \(x\)? If it is true, then find the values of \(x\).</p>
<p>True for all \(x\)</p>
<p>True for \(x \in (0, \pi/4)\)</p>
<p>True for \(x \in (\pi/4, \pi/2)\)</p>
<p>Not true for any value of \(x\)</p>
Step-by-Step Solution
Key Concept: The expression |tan x + cot x| simplifies to |2/sin(2x)| using algebraic manipulation and double angle formulas. This is always ≥ 2 because |sin(2x)| ≤ 1, making the inequality impossible to satisfy for any real x.
<p><strong>Step 1:</strong> Simplify |tan x + cot x|</p><p>tan x + cot x = sin x/cos x + cos x/sin x = (sin²x + cos²x)/(sin x cos x) = 1/(sin x cos x) = 2/sin(2x)</p><p>Therefore, |tan x + cot x| = |2/sin(2x)|</p><p><strong>Step 2:</strong> Analyze the range</p><p>For the expression to be defined, sin(2x) ≠ 0, which means x ≠ nπ/2 for any integer n.</p><p>Since |sin(2x)| ≤ 1 for all x where it's defined, we have |2/sin(2x)| ≥ 2</p><p><strong>Step 3:</strong> Conclusion</p><p>The inequality |tan x + cot x| < 1 requires |2/sin(2x)| < 1, which means |sin(2x)| > 2.</p><p>Since |sin(2x)| ≤ 1 always, this is impossible.</p><p>∴ Answer: The inequality has <strong>no solution</strong> (D)</p>
Correct Answer: D