Question

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

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

Answer 1

Answer: (B)

$\frac{17}{6}$

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

Let ${\cos }^{-1}\frac{4}{5}=\theta$

$\Rightarrow \cos \theta =\frac{4}{5}$

$\Rightarrow \sin \theta =\frac{3}{5}$

$\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{3}{4}$

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

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

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