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Question
Prove that `(sin(4"A" - 2"B") + sin(4"B" - 2"A"))/(cos(4"A" - 2"B") + cos(4"B" - 2"A"))` = tan(A + B)
Solution
`(sin(4"A" - 2"B") + sin(4"B" - 2"A"))/(cos(4"A" - 2"B") + cos(4"B" - 2"A")) = (sin{(4"A" - 2"B" + 4"B" - 2"A")/2} cos {(4"A" - 2"B" - 4"B" + 2"A")/2})/(cos{(4"A" - 2"B" + 4"B" - 2"A")/2} cos{(4"A" - 2"B" - 4"B" + 2"A")/2})`
= `(sin((2"A" + 2"B")/2) * cos((6"A" - 6"B")/2))/(cos((2"A" + 2"B")/2) * cos((6"A" - 6"B")/2)`
= `(sin("A" + "B"))/(cos("A" + "B"))`
= tan(A + B)
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