Question

tan7Atan4Atan3A =tan7A-tan4A-tan3A

Answer 1

proved ...............................

Answer 2

$\tan 7A\tan 4A\tan 3A=\tan 7A-\tan 4A-\tan  3A$, proved.

Step-by-step explanation:

Take 4A + 3A = 7A  

Taking both sides tan, we get

∴  $\tan (4A + 3A) = \tan 7A$

⇒ $\dfrac{ \tan 4A + \tan  3A}{1- \tan 4A\tan  3A}=\tan 7A$ ..... (1)

[ ∵ $\tan (A+B) =\dfrac{ \tan A + \tan  B}{1- \tan A \tan B}$]

By cross-multiplication (1), we get

⇒$\tan 4A + \tan  3A=(1- \tan 4A\tan 3A)\tan 7A$

⇒$\tan 4A + \tan  3A=\tan 7A- \tan 7A\tan 4A\tan 3A$

⇒$\tan 7A\tan 4A\tan 3A=\tan 7A-\tan 4A-\tan  3A$, proved.