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Saturday, January 29, 2022

2022/013) How do you prove that \tan^{-1}\frac{1}{5}+ \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8} = \frac{\pi}{4}

Let us have \tan^{-1}\frac{1}{5} = \alpha

\tan^{-1}\frac{1}{7} = \beta

\tan^{-1}\frac{1}{3} = \gamma

\tan^{-1}\frac{1}{8} = \delta

we need to find \alpha + \beta + \gamma + \delta

Let us add 2 at a time \alpha + \beta , \gamma +\delta and then \alpha + \beta + \gamma + \delta

We have 

\tan \alpha = \frac{1}{5}\cdots(1)

and  \tan \beta = \frac{1}{7}\cdots(2)

So \tan (\alpha + \beta) = \frac{\tan (\alpha) + \tan (\beta)}{1 - \tan (\alpha) \tan (\beta)}

= \frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \frac{1}{7}}

= \frac{\frac{12}{35}}{ \frac{34}{35}} = \frac{12}{34} = \frac{6}{17} 

We have 

\tan \gamma = \frac{1}{3}\cdots(3)

and  \tan \delta  = \frac{1}{8}\cdots(4)

So \tan (\gamma + \delta) = \frac{\tan (\gamma) + \tan (\delta)}{1 - \tan (\gamma) \tan (\delta)}

= \frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \frac{1}{8}}

= \frac{\frac{11}{24}}{ \frac{23}{24}} = \frac{11}{23} 

now \tan (\alpha + \beta + \gamma + \delta)

=\frac{\tan (\alpha + \beta)  + \tan (\gamma + \delta)}{1 - \tan (\alpha + beta) \tan (\gamma + \delta)}

= \frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \frac{17}{23}}

= \frac{\frac{6 * 23 + 11 * 17}{17 * 23}}{1-  \frac{66}{17 * 23}}

= \frac{6 * 23 + 11 * 17}{17 * 23 - 66}

=\frac{ 325}{325} = 1

so \tan (\alpha + \beta + \gamma + \delta) = \tan \frac{\pi}{4}

but we need to show that \alpha + \beta + \gamma + \delta = \frac{\pi}{4}

as it could be \frac{5\pi}{4}

but as  \alpha,\beta,\gamma, \delta each is less than \frac{\pi}{4} so sum is less than \pi so it is \frac{\pi}{4} 

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