Trig identies ..asap ! (1.)2 sin a sin b = a. cos(a-b)-cos(a+b) b. sin(a-b)+tan(a+b) c. sin (a+b)+cos(a-b) ( a - cos a)^2 a. 1-sin2a b. -1+cos a c. sec a +1

(1.)This relies on the angle-sum/difference identity
cos(a+-b)=cos(a)*cos(b)+-sin(a)*sin(b)

2 sin A sin B =
split product into sum:
sin(A)*sin(B)+sin(A)*sin(B)=
add cos(A)*cos(B)-cos(A)*cos(B)=0 -> doesn't break equality
sin(A)*sin(B)+sin(A)*sin(B)+cos(A)*cos(B)-cos(A)*cos(B)=
reorder a bit
cos(A)*cos(B)+sin(A)*sin(B)-cos(A)*cos(B)sin(A)*sin(B)=
add some [] to make the application of the above identity clearer, also note the flipping of +/- on the right one, because of the - in front of the []
[cos(A)*cos(B)+sin(A)*sin(B)] -[ cos(A)*cos(B)-sin(A)*sin(B)]=
A. cos(A-B)-cos(A+B)

and is also listed here (among lots of other identities):
en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities

(2.)This relies on the double angle identity
sin(2x)=2*sin(x)*cos(x)
 and half-angle identities
(1/2)[1-cos(2x)]=sin²(x)
(1/2)[1+cos(2x)]=cos²(x)

(sin A - cos A)^2=
sin²(A)-2*sin(A)*cos(A)+cos²(A)=
apply half angle identities
(1/2)[1-cos(2A)]-2*sin(A)*cos(A)+(1/2)[1+cos(2A)]=
(1/2)[1-cos(2A)+1+cos(2A)]-2*sin(A)*cos(A)=
(1/2)[2]-2*sin(A)*cos(A)=
1-2*sin(A)*cos(A)=
apply double angle identity
1-sin(2A)
A. 1-sin2A

It has a slope of 1 nd a y intercept of -2

2xy/9 is the answer to this problem