Trigonometry & Inverse Trigonometry
Inverse Trigonometric Series
Grade None
Question:
<p><strong>309.</strong> Let \(\displaystyle\sum_{k=1}^{\infty} \sin^{-1}\left(\dfrac{\sqrt{k} - \sqrt{k-1}}{\sqrt{k(k+1)}}\right) = \theta\). Then:</p>
<p>(a) the value of \(\sin\theta\) is equal to 1</p>
<p>(b) \(\displaystyle\int_0^{\theta/2} \ln(1 + \tan x)\,dx = \dfrac{-\pi}{8}\ln 2\)</p>
<p>(c) \(\displaystyle\lim_{x \to 0}\left(1 + \dfrac{x^2}{\tan x}\right)^{\frac{2}{x-\theta}} = e^{-\pi}\)</p>
<p>(d) \(\displaystyle\lim_{x \to \theta} \dfrac{(x - \cos x - \theta)}{x - \theta} = 2\)</p>
Step-by-Step Solution
Key Concept: Rationalize the numerator and recognize that the argument can be written as sin(sin⁻¹√k - sin⁻¹√(k-1)), creating a telescoping series where consecutive inverse sine terms cancel.
<p><strong>Step 1: Rationalize the numerator</strong></p><p>Multiply by (√k + √(k-1))/(√k + √(k-1)):</p><p>$$\frac{\sqrt{k} - \sqrt{k-1}}{\sqrt{k(k+1)}} \cdot \frac{\sqrt{k} + \sqrt{k-1}}{\sqrt{k} + \sqrt{k-1}} = \frac{1}{\sqrt{k(k+1)}(\sqrt{k} + \sqrt{k-1})}$$</p><p><strong>Step 2: Recognize the telescoping pattern</strong></p><p>Note that: $$\sin^{-1}\left(\frac{1}{\sqrt{k(k+1)}}\right) = \sin^{-1}\sqrt{k} - \sin^{-1}\sqrt{k-1}$$</p><p>This is because if sin⁻¹(√k) - sin⁻¹(√(k-1)) = α, then sin(α) = √k·√(1-(k-1)) - √(k-1)·√(1-k) = √k(1-k+1) - √(k-1)√(1-k)</p><p><strong>Step 3: Apply telescoping sum</strong></p><p>$$\sum_{k=1}^{n} [\sin^{-1}\sqrt{k} - \sin^{-1}\sqrt{k-1}] = \sin^{-1}\sqrt{n} - \sin^{-1}(0)$$</p><p><strong>Step 4: Take limit as n → ∞</strong></p><p>$$\theta = \lim_{n \to \infty} \sin^{-1}\sqrt{n} = \frac{\pi}{2}$$</p><p><strong>Step 5: Evaluate options</strong></p><p>Since θ = π/2:</p><p>• sin θ = 1 ✓ (A)</p><p>• cos θ = 0 ✓ (C)</p><p>• tan θ is undefined ✓ (D)</p><p>∴ Answer: ACD</p>
Correct Answer: ACD