Question

proove that : tan9A - tan6A -tan3A = tan9A.tan 6A.tan tan 3A​

Answer 1

SOLUTION

TO PROVE

$\displaystyle \sf{ \tan 9A - \tan 6A - { \tan 3A} = \tan 9A \tan 6A \tan 3A }$

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf{ \tan (A + B) = \frac{ \tan A + \tan B \: }{1 - \tan A \tan B } }$

PROOF

$\displaystyle \sf{ \tan 9A = \tan( 3A + 6A)}$

$\implies \displaystyle \sf{ \tan 9A = \frac{ \tan 3A + \tan 6A }{1 - \tan 3A \tan 6A } }$

On Cross Multiplication we get

$\displaystyle \sf{ \tan 9A - \tan 9A \tan 3A \tan 6A = { \tan 3A + \tan 6A } }$

$\implies \displaystyle \sf{ \tan 9A = { \tan 3A + \tan 6A } + \tan 9A \tan 3A \tan 6A }$

$\implies \displaystyle \sf{ \tan 9A - \tan 6A - { \tan 3A} = \tan 9A \tan 6A \tan 3A }$

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