$ an ^{-1} \frac{1}{3}+ an ^{-1} \frac{1}{5}+ an ^{-1} \frac{1}{7}+ an ^{-1} \frac{1}{8}$ has the value

Question

$	an ^{-1} \frac{1}{3}+	an ^{-1} \frac{1}{5}+	an ^{-1} \frac{1}{7}+	an ^{-1} \frac{1}{8}$ has the value

Aicharya · Accepted Answer

$\begin{aligned} & 	an ^{-1} \frac{1}{3}+	an ^{-1} \frac{1}{5}+	an ^{-1} \frac{1}{7}+	an ^{-1} \frac{1}{8} \ & =	an ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{3} 	imes \frac{1}{5}}ight)+	an ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7} 	imes \frac{1}{8}}ight) \ & =	an ^{-1}\left(\frac{4}{7}ight)+	an ^{-1}\left(\frac{3}{11}ight) \ & =	an ^{-1}\left(\frac{\frac{4}{7}+\frac{3}{11}}{1-\frac{4}{7} 	imes \frac{3}{11}}ight)\end{aligned}$ $=	an ^{-1}\left(\frac{65}{65}ight)=	an ^{-1}(1)=\frac{\pi}{4}$

$\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}$ has the value

Solution