Prove that: `(sin3A+sin5A +sin7A+sin9A)/(cos3A+cos5A+cos7A+cos9A)=tan6A`

Explanation

We have to proof,

`(sin3A+sin5A +sin7A+sin9A)/(cos3A+cos5A+cos7A+cos9A)=tan6A`

Let,

L.H.S=`(sin3A+sin5A +sin7A+sin9A)/(cos3A+cos5A+cos7A+cos9A)`

As we know the formula,
`sin(A^o)+sin(B^o)= 2sin(((A^o)+(B^o))/2)cos(((A^o)-(B^o))/2)`
and `(cos A+cosB)= 2cos(((A^o)-(B^o))/2)cos(((A^o)+(B^o))/2)`

then,

L.H.S.=`(2sin(((3A^o)+(5A^o))/2)cos(((3A^o)-(5A^o))/2)+2sin(((7A^o)+(9A^o))/2)cos(((7A^o)-(9A^o))/2))/(2cos(((3A^o)-(5A^o))/2)cos(((3A^o)+(5A^o))/2)+2cos(((7A^o)-(9A^o))/2)cos(((7A^o)+(9A^o))/2)`

L.H.S.=`(2sin(6A^o)cos(3A^o)+2sin(6A^o)cos(A^o))/(2cos(6A^o)cos(3A^o)+2cos(6A^o)cos(A^o))`

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