If A= arcsin(-4/5) in Quadrant IV and B=arctan(5/12) in Quadrant III, what is the values of tan(arcsin(-4/5)+arctan(5/12))? ASAP please...

-33/56

Step-by-step explanation:

suppose: A,B are the 2 angles of a triangle

we have: A = arcsin

               B = arctan

=> sin A = => cos A = (because A is in quadrant IV)

    tan B =

have:

because B is in quadrant III =>

tan(arcsin(-4/5)+arctan(5/12)) = tan( A + B)

but A,B are the 2 angles of a triangle => tan(A + B) =

have: sin(A+B) = sinA.cosB + cosA.sinB =

cos(A + B) = cosA.cosB - sinA.sinB =

=> tan(A + B) =

actually your concept is wrong

step-by-step explanation:

150% of x =300

150x=300×100

x=(300×100)÷150

x=200