Concept:

\({\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right) = {\tan ^{ - 1}}\left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right)\)

\({\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right)\)

Calculation:

As we know, \({\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right) = {\tan ^{ - 1}}\left( {\frac{x}{{\sqrt {1 - {x^2}} }}} \right)\)

⇒ \({\cos ^{ - 1}}\frac{4}{5} = {\tan ^{ - 1}}\frac{3}{4}\;\ and\ {\cos ^{ - 1}}\frac{{15}}{{17}} = {\tan ^{ - 1}}\frac{8}{{15}}\)

The expression \({\cos ^{ - 1}}\frac{4}{5} - {\cos ^{ - 1}}\frac{{15}}{{17}}\)  can be rewritten as follows:

\({\cos ^{ - 1}}\frac{4}{5} - {\cos ^{ - 1}}\frac{{15}}{{17}} = {\tan ^{ - 1}}\frac{3}{4} - {\tan ^{ - 1}}\frac{8}{{15}}\)

As we know that, \({\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x - y}}{{1 + xy}}} \right)\)

\({\tan ^{ - 1}}\frac{3}{4} - {\tan ^{ - 1}}\frac{8}{{15}} = {\tan ^{ - 1}}\left( {\frac{{\frac{3}{4} - \frac{8}{{15}}}}{{1 + \frac{2}{5}}}} \right)\)

\(= {\tan ^{ - 1}}\left( {\frac{{13}}{{84}}} \right)\)