Question

tan7A - tan4A - tan3A = tan7A.tan4A.tan3A​

Answer 1

Step-by-step explanation:

We have,

$\rm{7A = 4A + 3A}$

$\rm{ \implies \: tan(7A) = tan(4A + 3A)}$

We know,

$\boxed{ \tan( \alpha + \beta ) = \dfrac{ \tan( \alpha ) + \tan( \beta ) }{1 - \tan( \alpha ) \tan( \beta ) } }$

So,

$\rm{ \implies \: tan(7A) = \dfrac{ tan(4A) + tan(3A)}{1 - tan(4A) \cdot \: tan(3A)}}$

$\rm{ \implies \: tan(7A) \left( 1 - tan(4A) \cdot \: tan(3A)\right)= tan(4A) + tan(3A)}$

$\rm{ \implies \: tan(7A) - tan(4A) \cdot \: tan(3A)\cdot \: tan(7A)= tan(4A) + tan(3A)}$

$\rm{ \implies \: tan(7A) = tan(4A) + tan(3A) + tan(4A) \cdot \: tan(3A)\cdot \: tan(7A)}$

$\rm{ \implies \: tan(7A) - tan(4A) - tan(3A) = tan(4A) \cdot \: tan(3A)\cdot \: tan(7A)}$