Question

$\tan [{\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3}]=$

• A. $\frac{6}{17}$
• B. $\frac{17}{6}$
• C. $\frac{7}{16}$
• D. $\frac{16}{7}$

Answer 1

Answer: (B)

$\frac{17}{6}$

Given, $\tan \left[{\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3}\right]$
$=\tan [{\tan }^{-1}\frac{\sqrt{1-{\left(\frac{4}{5}\right)}^2}}{\frac{4}{5}}+{\tan }^{-1}\frac{2}{3}]$
$=\tan \left[{\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{2}{3}\right]=\tan \left[{\tan }^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times \frac{2}{3}}\right)\right]$
$=\tan \left[{\tan }^{-1}\left(\frac{9+8}{12-6}\right)\right]=\tan \left[{\tan }^{-1}\left(\frac{17}{6}\right)\right]$

$=\frac{17}{6}$