Question

lim x→0 [sin(a+b)x+sin(a-b)x+sin2ab]/cos²bx-cos²ax

Answer 1

sin((a + b) * x) + sin((a - b) * x) + sin(2ax) =>

sin(ax)cos(bx) + sin(bx)cos(ax) + sin(ax)cos(bx) - sin(bx)cos(ax) + sin(2ax) =>

2 * sin(ax) * cos(bx) + sin(2ax) =>

2 * sin(ax) * cos(bx) + 2 * sin(ax) * cos(ax) =>

2 * sin(ax) * (cos(bx) + cos(ax))

cos(bx)^2 - cos(ax)^2 =>

(cos(bx) - cos(ax)) * (cos(bx) + cos(ax))

2 * sin(ax) / (cos(bx) - cos(ax))

I don't know what x is going to, but that's simplified for now.

Answer 2

Answer:

Step-by-step explanation:  

                        The answer is (2a^2)/(a^2-b^2)