Question

sin2x+sin2y/sin2x-sin2y=tan(x+y)/tan(x-y) ​

Answer 1

Step-by-step explanation:

Prove that:

$\dfrac{sin2x+sin2y}{sin2x-sin2y}\ =\ \dfrac{tan(x+y)}{tan(x-y)}$

Following are the formulas which are used in this question:

$SinC+SinD\ =\ 2Sin\dfrac{C+D}{2}Cos\dfrac{C-D}{2}$

$SinC-SinD\ =\ 2Cos\dfrac{C+D}{2}Sin\dfrac{C-D}{2}$

For prove equal ,first we take L.H.S.

$=\dfrac{sin2x+sin2y}{sin2x-sin2y}$

Applying above formulas ,We get:

$=\dfrac{2Sin\dfrac{2x+2y}{2}Cos\dfrac{2x-2y}{2}}{2Cos\dfrac{2x+2y}{2}Sin\dfrac{2x-2y}{2}}$

After solving the numerator and the  denominator , we get:

$=\ \dfrac{Sin(x+y)Cos(x-y)}{Cos(x+y)Sin(x-y)}$

$=\ \dfrac{tan(x+y)}{tan(x-y)}$

L.H.S. =  R.H.S.

Above are the steps of proving the left hand side is equal to the right hand side.