Let #arc sin(3/5)=x, and, arc tan(-2)=y; x,y in (-pi/2,pi/2).#
#:. sinx=3/5, and, tany=-2.#
We note that, #sinx > 0, x in (0,pi/2), &, tany <0, y in (-pi/2,0).#
Since, the Reqd. Value
#=sin(x+y)=sinxcosy+cosxsiny,.....(ast)#
We have to find the values of #cosy, siny and cosx.#
#sinx=3/5 rArr cosx=sqrt(1-sin^2x)=sqrt(1-9/25)=+-4/5,#
#"but, "because, x in (0,pi/2), cosx=+4/5.........(1).#
Again, #tany=-2 rArr secy=sqrt(1+tan^2y)=+-sqrt5,# but with
#y in (-pi/2,0), secy=+sqrt5," giving, "cosy=+1/sqrt5...(2).#
Finally, #siny=tany*cosy=-2*1/sqrt5=-2/sqrt5...........(3).#
Utilising #(1),(2) and (3) in (ast),# we get,
#"The Reqd. Value="(3/5)(1/sqrt5)+(4/5)(-2/sqrt5)=-1/sqrt5.#
Enjoy Maths.!
