सिद्ध कीजिए: $\tan ^{-1} \frac{1}{5}$ + $\tan ^{-1} \frac{1}{7}$ + $\tan ^{-1} \frac{1}{3}$ + $\tan ^{-1} \frac{1}{8}$ = $\frac{\pi}{4}$
Miscellaneous Exercise-8
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दिया है, $\tan ^{-1} \frac{1}{5}$ + $\tan ^{-1} \frac{1}{7}$ + $\tan ^{-1} \frac{1}{3}$+ $\tan ^{-1} \frac{1}{8}$ = $\frac{\pi}{4}$
बायाँ पक्ष = $\left(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}\right)$ + $\left(\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}\right)$
= $\tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5} \times \frac{1}{7}}\right)$ + $\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3} \times \frac{1}{8}}\right)$ [$\because \tan ^{-1} x+\tan ^{-1} y$$=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$]
= $\tan ^{-1}\left(\frac{\frac{7+5}{35}}{\frac{35-1}{35}}\right)$+ $\tan ^{-1}\left(\frac{\frac{8+3}{24}}{\frac{24-1}{24}}\right)$= $\tan ^{-1}\left(\frac{6}{17}\right)$ + $\tan ^{-1}\left(\frac{11}{23}\right)$
= $\tan ^{-1}\left(\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17} \times \frac{11}{23}}\right)$ = $\tan ^{-1}\left(\frac{6 \times 23+11 \times 17}{17 \times 23-6 \times 11}\right)$
= $\tan ^{-1}\left(\frac{325}{325}\right)$ = $\tan ^{-1} 1$ = $\frac{\pi}{4}$ = दायाँ पक्ष इति सिद्धम्
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