Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions - Summation
Grade 12
Question:
<p>If \(\sum_{r=1}^{100} \sin^{-1}\left(\dfrac{1}{\sqrt{r^2+1}\sqrt{r^2+2r+2}}\right)\) is equal to \(\tan^{-1}\left(\dfrac{p}{q}\right)\) where \(p\) and \(q\) are co-prime, then the value of \((p+q)\) is equal to:</p>
<p>(a) 99</p>
<p>(b) 100</p>
<p>(c) 101</p>
<p>(d) 102</p>
Step-by-Step Solution
Key Concept: Recognize that the general term can be decomposed as a telescoping series using the identity sin⁻¹(x) = tan⁻¹(x/√(1-x²)), or equivalently, express the argument as a difference of inverse tangents: tan⁻¹(r+1) - tan⁻¹(r).
<p><strong>Step 1:</strong> Simplify the argument of sin⁻¹.</p><p>Note that r² + 2r + 2 = (r+1)² + 1</p><p>So the expression becomes: sin⁻¹(1/√[(r²+1)((r+1)²+1)])</p><p><strong>Step 2:</strong> Use the identity for inverse tangent differences.</p><p>Recall that tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab)) when ab > -1</p><p>We can verify that: sin⁻¹(1/√[(r²+1)((r+1)²+1)]) = tan⁻¹(r+1) - tan⁻¹(r)</p><p><strong>Step 3:</strong> Sum the telescoping series.</p><p>∑(r=1 to 100) [tan⁻¹(r+1) - tan⁻¹(r)] = tan⁻¹(101) - tan⁻¹(1)</p><p><strong>Step 4:</strong> Apply the tangent difference formula.</p><p>tan⁻¹(101) - tan⁻¹(1) = tan⁻¹((101-1)/(1+101·1)) = tan⁻¹(100/102) = tan⁻¹(50/51)</p><p><strong>Step 5:</strong> Identify p and q.</p><p>Since gcd(50, 51) = 1, we have p = 50 and q = 51</p><p>∴ Answer: p + q = 50 + 51 = <strong>101</strong></p>
Correct Answer: D