Question

Find the value of expression 2√sinAcos(90-A)+3√cosAsin(90-A)÷6sin(90-A)+4cos(90-A)

Answer 1

Answer:

We have, (2√sinAcos(90-A)+3√cosAsin(90-A))/(6sin(90-A)+4cos(90-A))

= (2√sin^2A+3√cos^2A)/(6cosA+4sinA)

= (2sinA+3cosA)/2.(3cosA+2sinA)

= 1/2