Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>\(\cot^{-1}\sqrt{\frac{1+x^2}{1-x^2}}\) is equal to</p>
<p>(a) \(\cos^{-1}(x^2)\)</p>
<p>(b) \(\frac{1}{2}\cos^{-1}(x^2)\)</p>
<p>(c) \(\frac{\pi}{2}\cos^{-1}(x^2)\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Convert the inverse cotangent expression using the relationship cot⁻¹(y) = tan⁻¹(1/y), then apply the double angle formula for cosine to simplify the argument.
<p><strong>Step 1:</strong> Let y = cot⁻¹√[(1+x²)/(1-x²)]. Then cot(y) = √[(1+x²)/(1-x²)]</p><p><strong>Step 2:</strong> Using cot(y) = 1/tan(y), we have tan(y) = 1/√[(1+x²)/(1-x²)] = √[(1-x²)/(1+x²)]</p><p><strong>Step 3:</strong> Recognize that if tan(y) = √[(1-x²)/(1+x²)], let's use the substitution x = cos(θ). Then tan(y) = √[(1-cos²θ)/(1+cos²θ)] = √[sin²θ/(1+cos²θ)]</p><p><strong>Step 4:</strong> Actually, use the cosine double angle approach: Note that (1-x²)/(1+x²) = (1-x²)/(1+x²). If we set x² = cos(α), then (1-cos α)/(1+cos α) relates to half-angle formulas.</p><p><strong>Step 5:</strong> More directly: Let tan(y) = √[(1-x²)/(1+x²)]. Using the identity tan(y/2) = √[(1-cos(2y))/(1+cos(2y))], if tan(y) = √[(1-x²)/(1+x²)], then comparing: cos(2y) = x². Therefore 2y = cos⁻¹(x²), giving y = (1/2)cos⁻¹(x²)</p><p><strong>Step 6:</strong> Thus cot⁻¹√[(1+x²)/(1-x²)] = (1/2)cos⁻¹(x²)</p><p><strong>∴ Answer:</strong> B</p>
Correct Answer: B