Question

If $\alpha ={\cos }^{-1}\left(\frac{3}{5}\right),\beta ={\tan }^{-1}\left(\frac{1}{3}\right),$ where $0<\alpha ,\beta <\frac{\pi }{2},$ then $\alpha -\beta$ is equal to:

• A. ${\tan }^{-1}\left(\frac{9}{14}\right)$
• B. ${\sin }^{-1}\left(\frac{9}{5\sqrt{10}}\right)$
• C. ${\cos }^{-1}\left(\frac{9}{5\sqrt{10}}\right)$
• D. ${\tan }^{-1}\left(\frac{9}{5\sqrt{10}}\right)$

Answer 1

The correct option is B ${\sin }^{-1}\left(\frac{9}{5\sqrt{10}}\right)$

Given,
$\alpha ={\cos }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{4}{3}\right)$
$\beta ={\tan }^{-1}\left(\frac{1}{3}\right)$

Now,
$\Rightarrow \alpha -\beta ={\tan }^{-1}\left(\frac{4}{3}\right)-{\tan }^{-1}\left(\frac{1}{3}\right)$
$={\tan }^{-1}\left(\frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{3}\cdot \frac{1}{3}}\right)$

$={\tan }^{-1}\left(\frac{9}{13}\right)$

$={\sin }^{-1}\left(\frac{9}{\sqrt{250}}\right)$

$={\sin }^{-1}\left(\frac{9}{5\sqrt{10}}\right)$