We have #tan(u+v)=(tanu+tanv)/(1-tanutanv)#, so we need #tanu# and #tanv# for finding #tan(u+v)#.
As #/_u# is in #Q1#, all trigonometric ratios are positive. As #cosu=4/7#, #secu=1/(4/7)=7/4# and using #tan^u=sec^2u-1#,
#tanu=sqrt(sec^2u-1)=sqrt((7/4)^2-1)#
= #sqrt(49/16-1)=sqrt(33/16)=sqrt33/4#
and #/_v# is #Q3#, while #tanv# and #cotv# are positive, all other trigonometric ratios are negative.
As #sinv=-9/10#, #cos^2v=1-sin^2v=1-(-9/10)^2=1-81/100=19/100#
and #cosv=-sqrt19/10#
and #tanv=sinv/cosv=(-9/10)/(-sqrt19/10)=9/sqrt19#
Hence #tan(u+v)=(tanu+tanv)/(1-tanutanv)#
= #(sqrt33/4-9/sqrt19)/(1-(sqrt33/4)xx(9/sqrt19)#
and multiplying numerator and denominator by #4sqrt19#, we have
#tan(u+v)=(sqrt19xxsqrt33+9xx4)/(4sqrt19-9sqrt33)#
= #(4.359xx5.745+36)/(4xx4.359-9xx5.745)#
= #(25.042455+36)/(17.436-51.705)#
= #-61.042455/34.269#
= #-1.7813#
