Step-by-step explanation:
64 bro
2/3 = x/(3x + y)
2*(3x + y) = 3*x
2*3x + 2*y = 3x
6x + 2y = 3x
6x - 3x = - 2y
3x = -2y
x/y = -2/3
x : y = -2/3.
First we will evaluate: ( substitution: u = x² - y², du = - 2 y dy )
=
( than plug in x and 0 )
=
=
= 1/3 x³ ( then another integration )
= 1/3 * 1/4 =
1/12
2-7x +10 =6
2-7(-2)+10
2-14+10
-16+10
-6
PEMDAS
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)