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Sin(A+B)Sin(A-B)

Prove that: t...

Question

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

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Solution

Determine the proof of the given expression $\tan 3 A \tan 2 A \tan A = \tan 3 A - \tan 2 A - \tan A$
Solve the L.H.S part:
Since we know that $3 A = 2 A + A . . . . . . . . . . . . . . . . . . . . . . . . . . . . (i)$
taking the $\tan$ both sides of the above equation.
$\tan 3 A = \tan (2 A + A) \Rightarrow \tan 3 A = \frac{\tan 2 A + \tan A}{(1 - \tan 2 A . \tan A)} \Rightarrow \tan 3 A (1 - \tan 2 A \tan A) = \tan 2 A + \tan A \Rightarrow \tan 3 A - \tan 3 A \tan 2 A \tan A = \tan 2 A + \tan A \Rightarrow \tan 3 A - \tan 2 A - \tan A = \tan 3 A \tan 2 A \tan A$
Hence, the L.H.S = R.H.S.

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