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Question

Prove:$$\tan 3A-\tan 2A-\tan A=\tan A\tan 2A\tan 3A$$.


Solution

$$LHS=\tan 3 A-\tan 2 A-\tan A=\tan (2 A+A)-(\tan 2 A+\tan A)=\dfrac{\tan 2 A+\tan A}{1-\tan 2 A\tan A}-(\tan 2 A+\tan A) $$
                                                            $$=(\tan 2 A+\tan A)\times \dfrac{1-1+\tan 2 A\tan A}{1-\tan 2 A\tan A}=\dfrac{\tan 2 A+\tan A}{1-\tan 2 A\tan A}\times \tan 2 A\tan A=\tan 3 A\tan 2 A\tan A=RHS$$
Hence Proved

Mathematics

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