Prove that: `(sin11 A\ sin A+sin7A\ sin3A)/(cos 11\ A\ sin A+cos7Asin3A)=t a n8A`

Explanation

We have to proof,

`(sin11 A\ sin A+sin7A\ sin3A)/(cos 11\ A\ sin A+cos7Asin3A)=tan8A`

Let,

L.H.S=`(sin11 A\ sin A+sin7A\ sin3A)/(cos 11\ A\ sin A+cos7Asin3A)`

Now divide and multiply by 2 we get,

L.H.S=`(2sin11 A\ sin A+2sin7A\ sin3A)/(2cos 11\ A\ sin A+2cos7Asin3A)`

As we know the formula,
`2sinAsinB=cos(A+B)-cos(A−B)` and
`2cosAsinB=sin(A+B)-sin(A−B)`

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