Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>The system of equations <span>\(\tan x = a \cot x\)</span>, <span>\(\tan 2x = b\cos y\)</span></p><p>Which of the following is/are correct?</p>
<p>(a) Cannot have a solution if <span>\(a = 0\)</span></p>
<p>(b) Cannot have a solution if <span>\(a = 1\)</span></p>
<p>(c) Cannot have a solution if <span>\(2 > |a| > |1-a|\)</span></p>
<p>(d) has a solution for all <span>\(a\)</span> and <span>\(b\)</span></p>
Step-by-Step Solution
Key Concept: Analyze when the system of trigonometric equations cannot have a solution by examining domain restrictions and solvability conditions.
<p><strong>Step 1:</strong> If <span>\(a = 0\)</span>, then <span>\(\tan x = 0 \Rightarrow x = n\pi\)</span> and for any value of <span>\(y\)</span> such that <span>\(\cos y = 0\)</span> the second equation satisfies. So option (a) is false.</p><p><strong>Step 2:</strong> If <span>\(a = 1\)</span> then <span>\(\tan x = \cot x \Rightarrow \tan 2x = 1\)</span>, but then <span>\(\tan 2x\)</span> is not defined. So option (b) is true.</p><p><strong>Step 3:</strong> From the first equation <span>\(\tan x = 4a\)</span>, so <span>\(a\)</span> must be positive. We have <span>\(|\cos y| = \frac{2\tan x}{b(1-\tan^2 x)} = \frac{24a}{b(1-a)} < 1\)</span>. Thus <span>\(24a < |b(1-a)|\)</span>, so option (c) is true.</p><p>∴ Answer is B,C.</p>
Correct Answer: B,C