Question

सिद्ध कीजिए: $\dfrac {(\sin 7x + \sin 5x ) + (\sin 9x + \sin 3x )}{(\cos 7x + \cos 5x ) + (\cos 9x + \cos 3x )} = \tan 6x$

Answer 1

Answer:

Step-by-step explanation:

सिद्ध  करना है -

$\dfrac {(\sin 7x + \sin 5x ) + (\sin 9x + \sin 3x )}{(\cos 7x + \cos 5x ) + (\cos 9x + \cos 3x )} = \tan 6x$

L.H.S. =$\dfrac {(\sin 7x + \sin 5x ) + (\sin 9x + \sin 3x )}{(\cos 7x + \cos 5x ) + (\cos 9x + \cos 3x )}$

   =  $\frac{2sin\frac{7x+5x}{2}cos\frac{7x-5x}{2}+2sin\frac{9x+3x}{2} cos\frac{9x-3x}{2}   }{2cos\frac{7x+5x}{2}cos\frac{7x-5x}{2}+2cos\frac{9x+3x}{2} cos\frac{9x-3x}{2}   }$

∵

    $sinC+sinD={2sin\frac{C+D}{2}cos\frac{C-D}{2}}\\cosC+cosD={2cos\frac{C+D}{2}cos\frac{C-D}{2}}$

 

= $\frac{2sin6xcosx+2sin6xcos3x}{2cos6xcosx+2cos6xcos3x}$

= $\frac{2sin6x(cosx+xos3x)}{2cos6x(cosx+cos3x)}$

  =$\frac{sin6x}{cos6x}$

  = tan6x  = R.H.S.

अतः   L.H.S. = R.H.S.