Prove that: `(sin3A cos4A-sin Acos2A)/(sin4A sin A+cos6A cos A)=tan2A`

Explanation

We have to proof,

`(sin3A cos4A-sin Acos2A)/(sin4A sin A+cos6A cos A)=tan2A`

Let,

L.H.S=`(sin3A cos4A-sin Acos2A)/(sin4A sin A+cos6A cos A)`

Now divide and multiply by 2 we get,

L.H.S=`(2sin3A cos4A-2sinA\ cos 2A)/(2sin 4A\ sinA-2cos6A cosA)`

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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