Q

# Prove that tan inverse 1 / 5 + tan inverse 1 / 7 + tan inverse 1 / 3 + tan inverse 1 / 8 = pi / 4

Prove that

8. $\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} +\tan^{-1}\frac{1}{3} +\tan^{-1}\frac{1}{8} = \frac{\pi}{4}$

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Applying the formlua:

$\tan^{-1}x + \tan^{-1}y = \tan^{-1} \frac{x+y}{1-xy}$    on two parts.

we will have,

$=\tan^{-1}\left (\frac{\frac{1}{5}+ \frac{1}{7}}{1- \frac{1}{5}\times \frac{1}{7}} \right ) + \tan^{-1}\left (\frac{\frac{1}{3}+ \frac{1}{8}}{1- \frac{1}{3}\times \frac{1}{8}} \right )$

$= \tan^{-1} \left ( \frac{7+5}{35-1} \right ) + \tan^{-1} \left ( \frac{8+3}{24-1} \right )$

$= \tan^{-1} \left ( \frac{12}{34} \right ) + \tan^{-1} \left ( \frac{11}{23} \right )$

$= \tan^{-1} \left ( \frac{6}{17} \right ) + \tan^{-1} \left ( \frac{11}{23} \right )$

$= \tan^{-1}\left [ \frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17}\times\frac{11}{23}} \right ]$

$= \tan^{-1}\left [ \frac{325}{325} \right ] = \tan^{-1} 1$

$=\frac{\pi}{4}$

Hence it s equal to R.H.S

Proved.

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