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Question
Prove that:
tan 4A tan 3A tan A + tan 3A + tan A – tan 4A = 0
Solution
4A = 3A + A
tan 4A = tan (3A + A)
tan 4A = `(tan 3"A" + tan "A")/(1- tan 3"A" tan "A")`
on cross multiplication we get,
tan 3A + tan A = tan 4A (1 – tan 3A tan A) = tan 4A – tan 4A tan 3A tanA
i.e., tan 4A tan 3A tan A + tan 3A + tan A = tan 4A
(or) tan 4A tan 3A tan A + tan 3A + tan A – tan 4A = 0
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