Question

$\dfrac{cos2B-cos2A}{sin2B+sin2A}$​

Answer 1

Answer is in Attachment

thank u

Answer 2

Answer:

$\mathsf{\dfrac{(cos2B-cos2A)}{(sin2B-sin2A)}}$

$\mathsf{\implies \dfrac{2sin(\dfrac{(2B+2A)}{2})sin(\dfrac{(2B-2A)}{2})}{2sin(\dfrac{(2B+2A)}{2})cos(\dfrac{(2B-2A)}{2})}}$

$\mathsf{\implies \dfrac{2sin(A+B)sin(B-A)}{2sin(B+A))cos(B-A)}}$

$\mathsf{\implies \dfrac{sin(A-B)}{cos(A-B)}}$

$\mathsf{\implies tan(A-B)}$

Step-by-step explanation:

Use the formulas of trigonometric identities .

The formulas to use :

$\mathtt{cosX-cosY=2sin(\dfrac{X+Y}{2})sin(\dfrac{X-Y}{2})}$

$\mathtt{sinX+sinY=2sin(\dfrac{X+Y}{2})cos(\dfrac{X-Y}{2})}$