Q (6)  Prove that 

\small \frac{(\sin 7x + \sin 5x) + (\sin9x + \sin 3x)}{(\cos7x + \cos5x) + (\cos9x + \cos3x)} = \tan6x

Answers (1)

We know that 

        sinA + sinB = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
        and 
         cosA + cosB =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
 
 We use these two identities in our problem

        sin7x + sin5x  =  2\sin\frac{7x+5x}{2}\cos\frac{7x-5x}{2}    =       2\sin6x\cos x
         
        sin 9x + sin 3x = 2\sin\frac{9x+3x}{2}\cos\frac{9x-3x}{2}  =    2\sin6x\cos 3x

       cos 7x + cos5x = 2\cos\frac{7x+5x}{2}\cos\frac{7x-5x}{2}   =   2\cos6x\cos x

       cos 9x + cos3x = 2\cos\frac{9x+3x}{2}\cos\frac{9x-3x}{2}  =2\cos6x\cos 3x


        \small \frac{(\sin 7x + \sin 5x) + (\sin9x + \sin 3x)}{(\cos7x + \cos5x) + (\cos9x + \cos3x)}      =      \small \frac{(2\sin 6x\cos x) + (2\sin6x \cos3x)}{(2\cos6x cos x) + (2\cos6x cos3x)}     

                                                                                       =     \small \frac{2\sin6x(\cos x + \cos3x)}{2\cos6x (cos x + cos3x)} = \tan6x       = R.H.S.                       \small \left ( \because \frac{\sin x}{\cos x} = \tan x\right )

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