Question

निम्नलिखित को सिद्ध कीजिए: $\sin^2 {6x} - \sin^2 {4x} = \sin 2x\,\sin 10x$

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

$sin^{2}6x - sin^{2}4x  =  sin2x . sin10x$

L.H.S.  =  sin^{2}6x - sin^{2}4x

           =  ( sin6x + sin4x ) ( sin6x - sin4x )

                 [ ∵ a^{2} - b^{2}  = (a+b) (a-b) ]

         

         =   ( sin6x + sin4x ) ( sin6x - sin4x )

          $(sinA - sinB = 2cos\frac{A+B}{2} sin\frac{A-B}{2})$

         ($sinA + sinB =2sin\frac{A+B}{2} cos\frac{A-B}{2}$)

          = $2sin(\frac{6x+4x}{2}) cos(\frac{6x-4x}{2}). 2cos\frac{6x+4x}{2} sin\frac{6x-4x}{2}$

          =  ( 2 sin5x . cosx ) ( 2 cos5x . sinx )

          =  ( 2 sin5x cos5x ) ( 2 sinx cosx )

         =  sin10x . sin2x

          [ ∵ 2 sinx cosx = sin2x ]

          =  R.H.S.

L.H.S.  =  R.H.S.