Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>If <span>\( f(x) = 2\tan^{-1}x + \sin^{-1}\left(\dfrac{2x}{1+x^2}\right) \)</span>, <span>\( x > 1 \)</span>, then <span>\( f(t) \)</span> is equal to</p>
<p>\( \dfrac{\pi}{2} \)</p>
<p>\( \pi \)</p>
<p>\( 4\tan^{-1}(5) \)</p>
<p>\( \tan^{-1}\left(\dfrac{65}{156}\right) \)</p>
Step-by-Step Solution
Key Concept: Recognize that sin⁻¹(2x/(1+x²)) can be expressed as 2tan⁻¹(x) when x > 1, using the substitution x = tan(θ) and the double angle formula for sine. The critical insight is that for x > 1, we must carefully handle the range of inverse sine to get 2tan⁻¹(x) - π.
<p><strong>Step 1:</strong> Let x = tan(θ) where θ ∈ (π/4, π/2) since x > 1.</p><p><strong>Step 2:</strong> Then 2x/(1+x²) = 2tan(θ)/(1+tan²(θ)) = 2tan(θ)/sec²(θ) = 2sin(θ)cos(θ) = sin(2θ)</p><p><strong>Step 3:</strong> For x > 1, we have θ ∈ (π/4, π/2), so 2θ ∈ (π/2, π). In this range, sin⁻¹(sin(2θ)) = π - 2θ (not 2θ, because the range of sin⁻¹ is [-π/2, π/2]).</p><p><strong>Step 4:</strong> Therefore sin⁻¹(2x/(1+x²)) = π - 2tan⁻¹(x) for x > 1.</p><p><strong>Step 5:</strong> f(x) = 2tan⁻¹(x) + π - 2tan⁻¹(x) = π</p><p><strong>Step 6:</strong> Thus f(t) = π for all t > 1.</p><p>∴ Answer: B</p>
Correct Answer: B