Question

Evaluate:

(a) ${\sin }^{-1}\frac{4}{5}+2{\tan }^{-1}\frac{1}{3}=\frac{\pi }{2}$

(b) ${\tan }^{-1}\frac{1}{7}+2{\tan }^{-1}\frac{1}{3}=\frac{\pi }{4}$

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

Answer 1

(a)

${\sin }^{-1}\left(\frac{4}{5}\right)+2{\tan }^{-1}\left(\frac{1}{3}\right)$

$={\sin }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left[\frac{2\times \frac{1}{3}}{1-{\left(\frac{1}{3}\right)}^2}\right]$

$={\sin }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right)$

$={\sin }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)$

$={\tan }^{-1}\left[\frac{\frac{4}{5}}{\sqrt{1-{\left(\frac{4}{5}\right)}^2}}\right]+{\tan }^{-1}\left(\frac{3}{4}\right)$

$={\tan }^{-1}\left(\frac{4}{3}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)$

$={\tan }^{-1}\left(\frac{4}{3}\right)+{\cot }^{-1}\left(\frac{4}{3}\right)$

$=\frac{\pi }{2}$

(b)

${\tan }^{-1}\left(\frac{1}{7}\right)+2{\tan }^{-1}\left(\frac{1}{3}\right)$

$={\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{\frac{2}{3}}{1-\frac{1}{9}}\right)$

$={\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)$

$={\tan }^{-1}\left(\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\times \frac{3}{4}}\right)$

$={\tan }^{-1}\left(\frac{\frac{25}{28}}{\frac{25}{28}}\right)$

$={\tan }^{-1}1$

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

(c)

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

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

$={\tan }^{-1}\left(\frac{\frac{12}{35}}{\frac{34}{35}}\right)+{\tan }^{-1}\left(\frac{\frac{11}{24}}{\frac{23}{24}}\right)$

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

$={\tan }^{-1}\left(\frac{\frac{325}{391}}{\frac{325}{391}}\right)$

$={\tan }^{-1}1$

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