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cos(pi/4-x)cos(pi/4-y)-sin(pi/4-x)sin(pi...

`cos(pi/4-x)cos(pi/4-y)-sin(pi/4-x)sin(pi/4-y)=sin(x+y)`

Text Solution

Verified by Experts

We have,
LHS =`cos(pi/4-A) cos(pi/4-B)-sin(pi/4-A)sin(pi/4-B)`
`=cos{(pi/4-A)+(pi/4-B)}=cos{pi/2-(A+B)}`
`=sin(A+B)`= RHS
Hence proved.
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