Question

More easy trig

Answer 1

\[(2\sin^2A-2\sin^2B)\div (\sin2A-\sin2B) = \tan(A+B)\]

Answer 2

@ikram002p

Answer 3

@ganeshie8

Answer 4

@ParthKohli

Answer 5

As you learned earlier,\[\sin^2 A - \sin^2 B = \sin(A + B)\sin(A - B)\]So the numerator is \(2\sin(A+ B)\sin(A - B)\).

How about the denominator? Use the identity:\[\sin C -\sin D = 2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)\]

Answer 6

ahhh thanks!

Answer 7

Great. Did you get the answer?