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)

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