Question:

Step 1: Establish the telescoping identity for the inverse sine expression. We need to verify that: $$\sin^{-1}\left(\frac{1}{\sqrt{r^2+1}\sqrt{r^2+2r+2}}\right) = \tan^{-1}(r+1) - \tan^{-1}(r)$$ Using the tangent subtraction formula, we have: $$\tan^{-1}(r+1) - \tan^{-1}(r) = \tan^{-1}\left(\frac{(r+1)-r}{1+r(r+1)}\right) = \tan^{-1}\left(\frac{1}{1+r^2+r}\right)$$ Note that $r^2+2r+2 = (r+1)^2+1$, so the denominator becomes: $$\sqrt{r^2+1} \cdot \sqrt{(r+1)^2+1}$$ This confirms our identity is correct. Step 2: Apply the telescoping sum to evaluate the series. The sum becomes: $$\sum_{r=1}^{100}\left[\tan^{-1}(r+1)-\tan^{-1}(r)\right]$$ This is a telescoping series where consecutive terms cancel: $$= \left[\tan^{-1}(2)-\tan^{-1}(1)\right] + \left[\tan^{-1}(3)-\tan^{-1}(2)\right] + \cdots + \left[\tan^{-1}(101)-\tan^{-1}(100)\right]$$ $$= \tan^{-1}(101) - \tan^{-1}(1)$$ Step 3: Simplify using the tangent subtraction formula. Apply the tangent difference formula: $$\tan^{-1}(101) - \tan^{-1}(1) = \tan^{-1}\left(\frac{101-1}{1+101 \cdot 1}\right)$$ $$= \tan^{-1}\left(\frac{100}{102}\right)$$ Reduce the fraction by dividing both numerator and denominator by 2: $$= \tan^{-1}\left(\frac{50}{51}\right)$$ Step 4: Identify the coprime integers and compute their sum. From the result $\tan^{-1}\left(\frac{50}{51}\right)$, we have: - $p = 50$ - $q = 51$ Verify that $\gcd(50, 51) = 1$, confirming they are coprime. Step 5: Calculate the final answer. $$p + q = 50 + 51 = 101$$ The answer is **101**, which corresponds to **Option 3**.