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If cos(A+B+C)=cosAcosBcosC , then find t...

If `cos(A+B+C)=cosAcosBcosC ,` then find the value of `(8sin(B+C)sin(C+A)sin(A+B))/(sin2Asin2Bsin2C)`

Text Solution

Verified by Experts

We have to prove that
`cos A cos (B+C)-sinA sin (B+C)=cos A cos B cos`
`rArr cos A (cos (B+C)-cos B cos C )=sin A sin (B+C)`
`rArr sin(B+C)=-(cos A sin B sinC)/(sin A)`
Similarly
`sin (C+A)=(-cos B sin C sin A)/(sin B)`
and `sin (A+B)=(-cos C sin A sin B)/(sin C)`
`therefore sin(A+B)sin(B+C)sin(C+A)`
`=-(cos A sin B sin C)/(sin A)(cos B sin Csin)/(sin B)`
`(cos C sin A sin)/(sin C)`
`=-(1)/(8)sin 2A sin 2B sin 2C`
`rArr (8sin(B+C)sin(C+A)sin(A+B))/(sin 2A sin 2B sin 2C)=-1`
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