If tan (A – B) = x, then the value of x is ;fn tan (A – B) = x,

sin (A – B) = sinA . cosB – cosA . sinB

cos (A – B) = cosA . cosB + sinA . sinB

$\therefore \frac{\sin (A-B)}{\cos (A-B)}$

$=\frac{\sin A \cdot \cos B-\cos A \cdot \sin B}{\cos A \cdot \cos B+\sin A \cdot \sin B}$

$\Rightarrow \tan (A-B)$

$=\frac{\frac{\sin A \cdot \cos A}{\cos A \cdot \cos B}-\frac{\cos A \cdot \sin B}{\cos A \cdot \cos B}}{\frac{\cos A \cdot \cos B}{\cos A \cdot \cos B}+\frac{\sin A \cdot \sin B}{\cos A \cdot \cos B}}$

(Dividing numerator and denominator by cosA cosB)

$=\frac{\tan A-\tan B}{1+\tan A \cdot \tan B}$