Prove that: `(sin5A cos2A-sin6A\ cos A)/(sin A\ sin2A-cos2A cos3A)=t a n A`

Explanation

We have to proof,

`(sin5A cos2A-sin6A\ cos A)/(sin A\ sin2A-cos2A cos3A)=tan A`

Let,

L.H.S=`(sin5A cos2A-sin6A\ cos A)/(sin A\ sin2A-cos2A cos3A)`

Now divide and multiply by 2 we get,

L.H.S=`(2sin5A cos2A-2sin6A\ cos A)/(2sin A\ sin2A-2cos2A cos3A)`

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

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