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.!
