Question

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

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

Answer 1

Answer: (A)

$\cos\alpha = \frac{3}{5} , \tan\beta = \frac{1}{3}$
$ \Rightarrow \tan\alpha = \frac{4}{3} $
$ \Rightarrow \tan\left(\alpha-\beta\right) = \frac{\frac{4}{3} - \frac{1}{3}}{1 + \frac{4}{3} . \frac{1}{3}} = \frac{9}{13} $
$ \Rightarrow \sin\left(\alpha-\beta\right) = \frac{9}{5\sqrt{10}}$
$ \Rightarrow \alpha - \beta = \sin^{-1} \left(\frac{9}{5\sqrt{10}}\right) $