Math Series #11

$\begin{aligned} X & = 10\cot \left(\cot^{-1} 3 + \cot^{-1} 7 + \cot^{-1} 13 + \cot^{-1} 21 \right) \\ & = 10\cot \left(\cot^{-1} \frac {3\cdot 7 -1}{3+7} + \cot^{-1} \frac {13\cdot 21 -1}{13+21} \right) \\ & = 10 \cot \left(\cot^{-1} 2 + \cot^{-1} 8 \right) \\ & = 10 \cot \left(\cot^{-1} \frac {2\cdot 8 -1}{2+8} \right) \\ & = 10 \cot \left(\cot^{-1} \frac 32 \right) \\ & = 10 \cdot \frac 32 \\ & = \boxed{15} \end{aligned}$

We shall apply the Inverse Trigonometric Identity , $\cot ^{-1} x+\cot^{-1} y=\cot^{-1}\left(\frac{xy-1}{x+y}\right)$ Note, $\begin{aligned} {\color{#69047E}\cot ^{-1} 3+\cot^{-1} 7}&=\cot^{-1}\left(\frac{3\cdot 7-1}{3+7}\right) \\ &=\cot^{-1} 2 \\ \\ {\color{#20A900}\cot ^{-1} 13+\cot^{-1} 21}&=\cot^{-1}\left(\frac{13\cdot 21-1}{13+21}\right) \\ &=\cot ^{-1} 8 \\ \\ {\color{#0C6AC7}\cot ^{-1} 2+\cot^{-1} 8}&=\cot^{-1}\left(\frac{2\cdot 8-1}{2+8}\right) \\ &=\cot^{-1} \frac{3}{2} \end{aligned}$ So, we can simplify the given expression as, $\begin{aligned} 10\cot (\cot^{-1}3+\cot^{-1}7 +\cot^{-1} 13+\cot^{-1} 21) &=10 \cot\left[\left({\color{#69047E}\cot^{-1}3+\cot^{-1}7} \right)+\left({\color{#20A900}\cot^{-1} 13+\cot^{-1} 21} \right)\right] \\ &=10\cot \left[{\color{#0C6AC7}\cot^{-1} 2+\cot ^{-1} 8} \right] \\ &=10\cot\left[\cot^{-1} \frac{3}{2}\right] \\ &=10\cdot\frac {3}{2} \\ &=\boxed{15} \end{aligned}$