<p>If \(\tan^{-1} y : \tan^{-1} x = 4:1\), express \(y\) as an algebraic function of \(x\). Hence or otherwise prove that \(22\frac{1}{2}\) is a root of the equation \(x^4 + 1 = 6x^2\).</p>
Step-by-Step Solution
Key Concept: Use the given ratio of inverse tangents to set up an equation, then apply the tangent addition formula repeatedly to express tan(4α) in terms of tan(α), where tan⁻¹(y) = 4tan⁻¹(x).
<p><strong>Step 1: Set up the relationship from the given ratio</strong></p><p>Given: tan⁻¹(y) : tan⁻¹(x) = 4:1</p><p>This means: tan⁻¹(y) = 4·tan⁻¹(x)</p><p>Taking tangent of both sides: y = tan(4·tan⁻¹(x))</p><p>Let α = tan⁻¹(x), so tan(α) = x and we need to find tan(4α).</p><p><strong>Step 2: Find tan(2α) using the double angle formula</strong></p><p>tan(2α) = (2tan(α))/(1 - tan²(α)) = (2x)/(1 - x²)</p><p><strong>Step 3: Find tan(4α) using the double angle formula again</strong></p><p>tan(4α) = (2tan(2α))/(1 - tan²(2α))</p><p>Let u = tan(2α) = (2x)/(1 - x²)</p><p>tan(4α) = (2u)/(1 - u²) = (2 · (2x)/(1-x²))/(1 - ((2x)/(1-x²))²)</p><p><strong>Step 4: Simplify the numerator</strong></p><p>Numerator = (4x)/(1 - x²)</p><p><strong>Step 5: Simplify the denominator</strong></p><p>Denominator = 1 - (4x²)/(1-x²)²</p><p>= ((1-x²)² - 4x²)/(1-x²)²</p><p>= (1 - 2x² + x⁴ - 4x²)/(1-x²)²</p><p>= (1 - 6x² + x⁴)/(1-x²)²</p><p><strong>Step 6: Combine to find y</strong></p><p>y = tan(4α) = ((4x)/(1-x²))/((1 - 6x² + x⁴)/(1-x²)²)</p><p>= (4x)/(1-x²) · (1-x²)²/(1 - 6x² + x⁴)</p><p>= 4x(1-x²)/(1 + x⁴ - 6x²)</p><p><strong>Step 7: Prove that 22½ is a root of x⁴ + 1 = 6x²</strong></p><p>If 22½° is a root of tan⁻¹(y) = 4·tan⁻¹(x), then at x = tan(22½°), we have y = tan(90°) which is undefined.</p><p>For y to be undefined, the denominator must equal zero: 1 + x⁴ - 6x² = 0, or x⁴ - 6x² + 1 = 0</p><p>Since x = tan(22½°) and 4·tan⁻¹(tan(22½°)) = tan⁻¹(tan(90°)), the denominator must vanish.</p><p>Therefore, tan(22½°) satisfies x⁴ + 1 = 6x².</p><p>∴ Answer: y = (4x(1-x²))/(1+x⁴-6x²)</p>
Correct Answer: \(y = \frac{4x(1-x^2)}{1+x^4-6x^2}\)