Question:

Step 1: Establish the identity for the sum term. The identity $\sec(r-1)\sec r = \frac{\tan r - \tan(r-1)}{\sin 1}$ is derived as follows: $$ \tan r - \tan(r-1) = \frac{\sin r}{\cos r} - \frac{\sin(r-1)}{\cos(r-1)} $$ $$ = \frac{\sin r \cos(r-1) - \cos r \sin(r-1)}{\cos r \cos(r-1)} $$ Using the sine subtraction formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $$ = \frac{\sin(r-(r-1))}{\cos r \cos(r-1)} = \frac{\sin 1}{\cos r \cos(r-1)} $$ Since $\sec x = \frac{1}{\cos x}$, we have: $$ \tan r - \tan(r-1) = \sin 1 \cdot \sec(r-1)\sec r $$ Rearranging, we get the desired identity: $$ \sec(r-1)\sec r = \frac{\tan r - \tan(r-1)}{\sin 1} $$ Step 2: Evaluate the sum. Substitute the identity into the sum: $$ \sin 1 \cdot \sum_{r=1}^{n} \sec(r-1)\sec r = \sin 1 \cdot \sum_{r=1}^{n} \frac{\tan r - \tan(r-1)}{\sin 1} $$ $$ = \sum_{r=1}^{n}(\tan r - \tan(r-1)) $$ This is a telescoping sum: $$ = (\tan 1 - \tan 0) + (\tan 2 - \tan 1) + \dots + (\tan n - \tan(n-1)) $$ $$ = \tan n - \tan 0 $$ Since $\tan 0 = 0$: $$ = \tan n $$ Step 3: Express $S_n$ using the principal value of $\tan^{-1}$. From the definition of $S_n$ and the sum evaluation: $$ S_n = \tan^{-1}(\tan n) $$ The principal value of $\tan^{-1}(x)$ lies in the interval $(-\pi/2, \pi/2)$. Therefore, for any real number $n$, $\tan^{-1}(\tan n)$ is given by $n - k\pi$, where $k$ is an integer chosen such that $n - k\pi \in (-\pi/2, \pi/2)$. Step 4: Calculate $S_5$. For $n=5$, we need to find an integer $k$ such that $5 - k\pi \in (-\pi/2, \pi/2)$. We use the approximations $\pi \approx 3.14159$ and $\pi/2 \approx 1.5708$. If $k=0$, $5 \notin (-\pi/2, \pi/2)$. If $k=1$, $5 - \pi \approx 5 - 3.14159 = 1.85841$. This value is not in $(-\pi/2, \pi/2)$ because $1.85841 > 1.5708$. If $k=2$, $5 - 2\pi \approx 5 - 2(3.14159) = 5 - 6.28318 = -1.28318$. This value is in $(-\pi/2, \pi/2)$ because $-1.5708 < -1.28318 < 1.5708$. Therefore, $S_5 = 5 - 2\pi$. Step 5: Calculate $S_{10}$. For $n=10$, we need to find an integer $k$ such that $10 - k\pi \in (-\pi/2, \pi/2)$. $10 \approx 3.1831 \pi$. If $k=3$, $10 - 3\pi \approx 10 - 3(3.14159) = 10 - 9.42477 = 0.57523$. This value is in $(-\pi/2, \pi/2)$ because $-1.5708 < 0.57523 < 1.5708$. Therefore, $S_{10} = 10 - 3\pi$.