Question

The value of $\tan [{\cos }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{2}{3}]$ is

• A. 18
• B. 2
• C. 9
• D. None of these

Answer 1

Answer: (A)

$\tan \left[{\cos }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{2}{5}\right]$
$=\tan \left[{\tan }^{-1}\frac{\sqrt{1-{\left(\frac{3}{5}\right)}^2}}{\frac{3}{5}}+{\tan }^{-1}\frac{2}{3}\right]$
$\left(\because {\cos }^{-1}x={\tan }^{-1}\frac{\sqrt{1-x^2}}{x}\right)$
$=\tan \left[{\tan }^{-1}\frac{\sqrt{25-9}}{5}\times \frac{5}{3}+{\tan }^{-1}\frac{2}{3}\right]$
$=\tan \left[{\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{2}{3}\right]$
${=\tan \left[{\tan }^{-1}\left\{\frac{\left(\frac{4}{3}+\frac{2}{3}\right)}{1-\left(\frac{4}{3}\right)\left(\frac{2}{3}\right)}\right\}\right]]}^{-1}$
$=\tan \left[{\tan }^{-1}\left\{\frac{6}{3}\times \frac{9}{1}\right\}\right]$
$=\tan \left[{\tan }^{-1}(18)\right]$
$=18$