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Question
Prove that:
`(cos 2"A" - cos 3"A")/(sin "2A" + sin "3A") = tan "A"/2`
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Solution
LHS = `(cos 2"A" - cos 3"A")/(sin "2A" + sin "3A")` ...`[∵ cos "C" - cos "D" = - 2 sin (("C + D")/2) sin (("C - D")/2)]`
`= (- 2 sin ((2"A" + 3"A")/2) sin((2"A" - 3"A")/2))/(2 sin ((2"A" + 3"A")/2) cos((2"A" - 3"A")/2))` ...`[∵ sin "C" + sin "D" = 2 sin (("C + D")/2) cos (("C - D")/2)]`
`= (- 2 sin((5"A")/2) sin ((- "A")/2))/(2 sin ((5"A")/2) cos ((- "A")/2))`
`= (2 sin ((5"A")/2) sin ("A"/2))/(2 sin ((5"A")/2) cos (("A")/2))`
= tan `("A"/2)` = RHS
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