Question

सिद्ध कीजिए: $(\cos x - \cos y)^{2} + (\sin x - \sin y)^{2} = 4\sin^2 \dfrac{x-y}{2}$

Answer 1

L.H.S =x+y=90 ,x=90-y ,y=90-x

=( cosx-cosy)^2+(sinx+siny)^2

=(cosx+sin(90-y))^2+(sinx+cos(90-y))^2

=(cosx+sinx)^2+(sinx+cosx)^2

=1+1

=2

Answer 2

Answer:

Step-by-step explanation:

सिद्ध करना है -

$(\cos x - \cos y)^{2} + (\sin x - \sin y)^{2} = 4\sin^2 \dfrac{x-y}{2}$

L.H.S.  =$(\cos x - \cos y)^{2} + (\sin x - \sin y)^{2}$

   

चूँकि हम जानते है कि -

     $cosA-cosB=-2sin\frac{A+B}{2} sin\frac{A-B}{2} \\sinA-sinB=2cos\frac{A+B}{2} sin\frac{A-B}{2}$

∴L.H.H.  =$[-2sin\frac{x+y}{2} sin\frac{x-y}{2}]^{2}+[2cos\frac{x+u}{2} sin\frac{x-y}{2}]^{2}$

          =$4sin^{2}\frac{x+y}{2} sin^{2}\frac{x-y}{2} +4cos^{2}\frac{x+y}{2} sin^{2}\frac{x-y}{2}$

        =$4sin^{2}\frac{x-y}{2} [sin^{2}\frac{x+y}{2}+ cos^{2}\frac{x+y}{2} ]$

       =$4sin^{2}\frac{x-y}{2} .1$

       =$4sin^{2}\frac{x-y}{2}$

       = R.H.S.