Question

$2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{5 \sqrt{2}}{7}+2 \tan ^{-1} \frac{1}{8}=$

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{8}$

Answer 1

Answer: (B)

$2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{5 \sqrt{2}}{7}+2 \tan ^{-} \frac{1}{8}$
$=2\left[\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}\right]+\sec ^{-1} \frac{5 \sqrt{2}}{7}$
$=2 \tan ^{-1} \frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{40}}+\sec ^{-1} \frac{5 \sqrt{2}}{7}$
$=2 \tan ^{-1} \frac{13}{39}+\sec ^{-1} \frac{5 \sqrt{2}}{7}$
$=2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7} \quad\left[\because \sec ^{-1} \frac{5 \sqrt{2}}{7}=\tan ^{-1} \frac{1}{7}\right]$
$=\tan ^{-1} \frac{\frac{2}{3}}{1-\frac{1}{9}}+\tan ^{-1} \frac{1}{7}$
$=\tan ^{-1} \frac{6}{8}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{\frac{3}{4}+\frac{1}{7}}{1-\left(\frac{3}{4} \times \frac{1}{7}\right)}$
$=\tan ^{-1}\left(\frac{21+4}{28-3}\right)=\tan ^{-1}(1)=\frac{\pi}{4}$