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Prove that : sin 51^0+cos 81^0=cos 21^0...

Prove that `: sin 51^0+cos 81^0=cos 21^0`

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We have to proof,
`sin(51^o)+cos(81^o)=cos(21^o)`
Let,
L.H.S.=`sin(51^o)+cos(81^o)`
L.H.S=`sin(51^o)+cos((90^o)-(9^o))`
L.H.S=`sin(51^o)-sin(9^o)`
As we know the formula,
`sin(A^o)+sin(B^o)= 2[sin(((A^o) +(B^o))/2)cos(((A^o)-(B^o))/2)]`
then, ...
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