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It cos(alpha+beta)=4/5,sin(alpha-beta)=5...

It `cos(alpha+beta)=4/5,sin(alpha-beta)=5/(13)a n dalpha,beta` lie between `0a n dpi/4` , prove that `tan2alpha=(56)/(33)`

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Given,
`cos(alpha+beta)=4/5`
`sin(alpha−beta)=5/13`
so,by pythagorous formula,
`sin(alpha+beta)=sqrt(1-(cos^2(alpha+beta)))`
`=>sqrt(1-(4/5)^2)`
=>`3/5`
Similarly we have,`cos(alpha-beta)=(12)/(13)`
...
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