Prove that: `(sinA+sin3A+sin5A+sin7A)/(cosA+cos3A+cos5A+cos7A)=tan4A`

Explanation

We have to proof,

`(sinA+sin3A+sin5A+sin7A)/(cosA+cos3A+cos5A+cos7A)=tan4A`

Let,

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

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(((A^o)+(3A^o))/2)cos(((A^o)-(3A^o))/2)+2sin(((5A^o)+(7A^o))/2)cos(((5A^o)-(7A^o))/2))/(2cos(((A^o)-(3A^o))/2)cos(((A^o)+(3A^o))/2)+2cos(((5A^o)-(7A^o))/2)cos(((5A^o)+(7A^o))/2)`

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

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