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Prove that: sin 80^0-cos 70^0=cos 50^0...

Prove that: `sin 80^0-cos 70^0=cos 50^0`

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We have to proof,
`sin(80^o)-cos(70^o)=cos(50^o)`
Let,
L.H.S.=`sin(80^o)-cos(70^o)`
L.H.S=`sin(80^o)-cos((90^o)-(20^o))`
L.H.S=`sin(80^o)-sin(20^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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