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(sin(B-C))/(cosBcosC)+sin(C-A)/(cosCcosA...

`(sin(B-C))/(cosBcosC)+sin(C-A)/(cosCcosA)+sin(A-B)/(cosAcosB)=0`

Text Solution

Verified by Experts

The first term of the LHS is
`(sin(B-C))/(cos B cos C)=(sin B cos C-cos B sin C)/(cos B cos C)`
`=(sin B cos C)/(cos B cos C)-(cos B sin C)/(cos B cos C)=tan B-tan C`
Similarly, the second term of the LHS is `(tan C-tan A)` and the third term of the LHS is `(tan A-tan B)`
Now LHS`=(tan B-tan C)+(tan C-tan A)`
`+(tan A-tan B)=0`
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