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Question
Prove that `(sin ("B - C"))/(cos "B" cos "C") + (sin ("C - A"))/(cos "C" cos "A") + (sin ("A - B"))/(cos "A" cos "B")` = 0
Solution
Consider `(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 ……… (1)
Similarly we can prove `(sin ("C - A"))/(cos "C" cos "A")` = tan C – tan A …….(2)
and `(sin ("A - B"))/(cos "A" cos "B")` = tan A – tan B …….. (3)
Add (1), (2) and (3) we get
`(sin ("B - C"))/(cos "B" cos "C") + (sin ("C - A"))/(cos "C" cos "A") + (sin ("A - B"))/(cos "A" cos "B")` = 0
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