Question

Prove that :
Cot ^-1 3 + Cot^-1 5 + Cot ^-17 + Cot^-1 8 = ​Π/ 4

Answer 1

$$\begin{gathered} \mathrm{LHS}={\cot }^{-1}3+{\cot }^{-1}5+{\cot }^{-1}7+{\cot }^{-1}8 \\ ={\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{8}\right) \\ =\left[{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)\right]+\left[{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{8}\right)\right] \\ ={\tan }^{-1}\left[\frac{{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{5}}}{1-{\displaystyle \frac{1}{3}}.{\displaystyle \frac{1}{5}}}\right]+{\tan }^{-1}\left[\frac{{\displaystyle \frac{1}{7}+\frac{1}{8}}}{1-{\displaystyle \frac{1}{7}}.{\displaystyle \frac{1}{8}}}\right] \\ ={\tan }^{-1}\left[\frac{8/15}{14/15}\right]+{\tan }^{-1}\left[\frac{15/56}{55/56}\right] \\ ={\tan }^{-1}\left(\frac{4}{7}\right)+{\tan }^{-1}\left(\frac{3}{11}\right) \\ ={\tan }^{-1}\left[\frac{{\displaystyle \frac{4}{7}+\frac{3}{11}}}{1-{\displaystyle \frac{4}{7}}.{\displaystyle \frac{3}{11}}}\right] \\ ={\tan }^{-1}\left[\frac{65/77}{65/77}\right] \\ ={\tan }^{-1}\left(1\right) \\ ={\tan }^{-1}\left[\tan \left(\frac{\mathrm{\pi }}{4}\right)\right] \\ =\frac{\mathrm{\pi }}{4} \\ =\mathrm{RHS} \end{gathered}$$