Question

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

Answer 1

Answer: 63

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