Question

If ${\cos }^{-1}\left(\frac{15}{17}\right)+2{\tan }^{-1}\left(\frac{1}{5}\right)={\cos }^{-1}\left(\frac{140}{k}\right)$ then find $k$

Answer 1

Answer: 221.00

${\cos }^{-1}\frac{15}{17}={\tan }^{-1}\frac{8}{15}$
$2{\tan }^{-1}\left(\frac{1}{15}\right)={\tan }^{-1}\left(\frac{5}{12}\right)$
$\therefore {\cos }^{-1}\left(\frac{15}{17}\right)+2{\tan }^{-1}\left(\frac{1}{5}\right)$
$={\tan }^{-1}\left(\frac{8}{15}\right)+{\tan }^{-1}\left(\frac{5}{12}\right)$
$={\tan }^{-1}\left(\frac{\frac{8}{15}+\frac{5}{12}}{1-\frac{40}{180}}\right)={\tan }^{-1}\left(\frac{171}{140}\right)$
$={\cos }^{-1}\left(\frac{140}{221}\right)$