Question

If $\alpha=cos^{-1} \left(\frac{3}{5}\right), \beta=tan^{-1} \left(\frac{1}{3}\right)$, where $0<\,\alpha, \,\beta <\,\frac{\pi}{2}$, and $\alpha-\beta$ is equal to $cos^{-1}\left(\frac{a}{b \sqrt{c}}\right)$, then $a + bc$ is

Answer 1

$\because cos\,\alpha = \frac{3}{5}$, then $sin\,\alpha =\frac{4}{5}$
$tan \,\alpha=\frac{4}{3}$
and $tan\,\beta = \frac{1}{3}$
$\because tan \left(\alpha-\beta\right)=\frac{tan\,\alpha-tan\,\beta}{1+tan\,\alpha.tan\,\beta} $
$=\frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{9}}=\frac{1}{\frac{13}{9}}=\frac{9}{13}
\therefore \alpha-\beta = tan^{-1}\left(\frac{9}{13}\right) $
$=sin^{-1} \left(\frac{9}{5\sqrt{10}}\right)=cos^{-1} 1\left(\frac{13}{5\sqrt{10}}\right)$
$\Rightarrow cos^{-1}\left(\frac{a}{b\sqrt{c}}\right)=cos^{-1}\left(\frac{13}{5\sqrt{10}}\right)$
$\Rightarrow a=13, b=5, c=10$
Hence $a + bc = 63$