Question

$\tan [co{s}^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3}]$ is equal to

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

Answer 1

The correct option is C $\frac{17}{6}$

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

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

$Here,\frac{3}{4}<\frac{2}{3}<1$

$=\tan \left[{\tan }^{-1}\frac{(\frac{3}{4}+\frac{2}{3})}{1-\frac{3}{4}\times \frac{2}{3}}\right]\dots (ta{n}^{-1}A+ta{n}^{-1}B=ta{n}^{-1}\frac{A+B}{1-AB})$

$=\tan \left[{\tan }^{-1}\frac{17}{6}\right]=\frac{17}{6}$