#"We can proceed as follows:" #
# sin( cos^{-1}(3/4) - tan^{-1}(1/3) ) #
# \qquad \qquad \qquad \qquad \qquad = \ sin( \overbrace{ cos^{-1}(3/4) }^{x} - \overbrace{ tan^{-1}(1/3) }^{y} ) #
# \qquad \ "now use:" \qquad \ sin ( x - y ) \ = \ sin(x)cos(y) - cos(x)sin(y) \qquad rArr #
# = \ sin ( \overbrace{ cos^{-1}(3/4) }^{x} ) cos( \overbrace{ tan^{-1}(1/3) }^{y} ) #
# \qquad \qquad \qquad \qquad \qquad - cos( \overbrace{ cos^{-1}(3/4) }^{x} ) sin( \overbrace{ tan^{-1}(1/3) }^{y} ) #
# "now use:" \quad sin ( x ) \ = \ sqrt{ 1 - cos^2 ( x ) }; \quad cos( cos^{-1} ( x ) ) \ = \ x \qquad rArr #
# = \ sqrt{ 1 - [ cos( \overbrace{ cos^{-1}(3/4) }^{x} ) ]^2 } cdot 1/{ csc( \overbrace{ tan^{-1}(1/3) }^{y} ) } #
# \qquad \qquad \qquad \qquad \qquad - \ (3/4) cdot 1/{ sec( \overbrace{ tan^{-1}(1/3) }^{y} ) } #
# = \ sqrt{ 1 - (3/4)^2 } cdot1/{ csc( \overbrace{ tan^{-1}(1/3) }^{y} ) } #
# \qquad \qquad \qquad \qquad \qquad - 3/4 cdot 1/{ sec( \overbrace{ tan^{-1}(1/3) }^{y} ) } #
# "now use:" \ \ csc( x ) = sqrt{ 1 + cot^2 ( x ) }; \ sec( x ) = sqrt{ 1 + tan^2 ( x ) } \ rArr #
# = \ sqrt{ 1 - 9/16 } cdot 1/sqrt{ 1 + [ cot( \overbrace{ tan^{-1}(1/3) }^{y} ) ]^2 #
# \qquad \qquad \qquad \qquad \qquad - 3/4 cdot 1/sqrt{ 1 + [ tan( \overbrace{ tan^{-1}(1/3) }^{y} ) ]^2 #
# = \ sqrt{ 1 - 9/16 } cdot 1/sqrt{ 1 + 1/[ tan( \overbrace{ tan^{-1}(1/3) }^{y} ) ]^2 #
# \qquad \qquad \qquad \qquad \qquad - 3/4 cdot 1/sqrt{ 1 + [ tan( \overbrace{ tan^{-1}(1/3) }^{y} ) ]^2 #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ "now use:" \qquad \quad tan( tan^{-1} ( x ) ) \ = \ x \qquad rArr #
# = \ sqrt{ 7/16 } cdot 1/sqrt{ 1 + [ 1/ (1/3)^2 ] } \ - \ 3/4 cdot 1/sqrt{ 1 + (1/3)^2 } #
# = \ sqrt{ 7 }/4 cdot 1/sqrt{ 1 + [ 1/{ 1/9 } ] } \ - \ 3/4 cdot 1/sqrt{ 1 + 1/9 } #
# = \ sqrt{ 7 }/4 cdot 1/sqrt{ 1 + 9 } \ - \ 3/4 cdot 1/sqrt{ 10/9 } #
# = \ sqrt{ 7 }/4 cdot 1/sqrt{ 10 } \ - \ 3/4 cdot sqrt{ 9/10 } #
# = \ sqrt{ 7 }/4 cdot 1/sqrt{ 10 } \ - \ 3/4 cdot 3/sqrt{ 10 } #
# = \ { sqrt{ 7} \ - \ 9 }/{ 4 sqrt{ 10 } } #
# = \ { ( sqrt{ 7} \ - \ 9 ) sqrt{ 10 } }/{ 4 cdot 10 } #
# = \ ( sqrt{ 70} \ - \ 9 sqrt{ 10 } )/{ 40 } \quad. #
# "This is our answer." #
# "Summarizing:" #
# \qquad \quad \ \ sin( cos^{-1}(3/4)- tan^{-1}(1/3) ) = \ ( sqrt{ 70} \ - \ 9 sqrt{ 10 } )/{ 40 } \quad. #
