Check the picture below.
make sure your calculator is in Degree mode.
Take L.H.S sin2A+sin2B/sin2A-sin2B
= sin2A+sin2B/sin2A-sin2B
Put
[sinC+sinD = 2sin(C+D)/2cos(C-D)/2]
[sinC-sinD = 2cos(C+D)/2.sin(C-D)/2]
= 2 sin(2A+2B)/2 cos(2A-2B)/2 / 2 cos(2A+2B) sin(2A-2B)
= sin(A+B).cos(A-B)/cos(A+B).sin(A-B)
= sin(A+B)/cos(A+B) . cos(A-B)/sin(A-B)
= tan(A+B).cot(A-B)
= tan(A+B).1/tan(A-B)
= tan(A+B)/tan(A-B)
∴ Hence we proved sin2A+sin2B/sin2A-sin2B=tan(A+B)/tan(A-B)
The y intercepts are when x=0 (the line crosses the y axis) and the x intercepts are when y=0 (the line crosses the x axis). Plugging x=0 in, we get 5*0=2y+4. Subtracting 4 from both sides, we get -4=2y and by dividing both sides by 2 we get y=-2, for one intercept to be (0, -2). For the other intercept, we make y=0 to get 5x=2*0+4=4. Dividing both sides by 4, we get x=4/5, making our other intercept (4/5, 0)
Answer:
sorry not sure
Step-by-step explanation:
