Question : If $\tan\left ( A-B \right )=x$, then the value of $x$ is:
Option 1: $\frac{\tan A+\tan B}{1-\tan A \tan B}$
Option 2: $\frac{\tan A+\tan B}{1+\tan A \tan B}$
Option 3: $\frac{\tan A-\tan B}{1-\tan A \tan B}$
Option 4: $\frac{\tan A-\tan B}{1+\tan A \tan B}$
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Correct Answer: $\frac{\tan A-\tan B}{1+\tan A \tan B}$
Solution :
$\tan(A-B) = \frac{\sin(A-B)}{\cos(A-B)}$
We know that,
$\sin(A-B)=\sin A \cos B - \cos A \sin B$
$\cos(A-B)=\cos A \cos B + \sin A \sin B$
On substituting the values,
$\tan(A-B) = \frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B + \sin A \sin B}$
Dividing the numerator and the denominator by $\cos A \cos B$, we get,
$⇒\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$
Hence, the correct answer is $\frac{\tan A - \tan B}{1 + \tan A \tan B}$.
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