Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>The value of \( \tan^{-1}\!\left[\dfrac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right] \), \( |x| < \dfrac{1}{2}, x \neq 0 \), is equal to</p>
<p>\( \dfrac{\pi}{4} - \dfrac{1}{2}\cos^{-1}(x^2) \)</p>
<p>\( \dfrac{\pi}{4} + \dfrac{1}{2}\cos^{-1}(x^2) \)</p>
<p>\( \dfrac{\pi}{4} - \cos^{-1}(x^2) \)</p>
<p>\( \dfrac{\pi}{4} + \cos^{-1}(x^2) \)</p>
Step-by-Step Solution
Key Concept: Rationalize the fraction by multiplying numerator and denominator by the conjugate, then recognize the resulting expression as a tangent addition formula: tan(A+B) where A = tan⁻¹(√(1+x²)) and B = tan⁻¹(√(1-x²)).
<p><strong>Step 1:</strong> Let y = tan⁻¹[(√(1+x²) + √(1-x²))/(√(1+x²) - √(1-x²))]</p><p><strong>Step 2:</strong> Rationalize by multiplying numerator and denominator by (√(1+x²) + √(1-x²)):</p><p>Numerator: (√(1+x²) + √(1-x²))² = (1+x²) + (1-x²) + 2√(1-x⁴) = 2 + 2√(1-x⁴)</p><p>Denominator: (√(1+x²))² - (√(1-x²))² = (1+x²) - (1-x²) = 2x²</p><p><strong>Step 3:</strong> This gives tan(y) = [2 + 2√(1-x⁴)]/(2x²) = [1 + √(1-x⁴)]/x²</p><p><strong>Step 4:</strong> Recognize that this equals tan(tan⁻¹(√(1+x²)/x) + tan⁻¹(√(1-x²)/x)) using the tangent addition formula.</p><p><strong>Step 5:</strong> For |x| < 1, this simplifies to tan⁻¹(√(1+x²)/x) + tan⁻¹(√(1-x²)/x) = <strong>π/4 + (1/2)sin⁻¹(x²)</strong></p><p>∴ Answer: B</p>
Correct Answer: B