If A and B are supplementary angles, then find the value of  $$\frac{\tan A + \tan B}{1 - \tan A \times \tan B}$$  ?

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Correct option is A

Given:

A and B are supplementary angles, which means A + B = 180°.

Formula Used:

We use the following identity for supplementary angles:
tan(180° - A) = -tan(A)
Also, if A and B are supplementary, then B = 180° - A.

Solution:

$$\frac{\tan A + \tan B}{1 - \tan A \times \tan B} \\\ \\A + B = 180^\circ \\\ \\A = (180^\circ - B) \\\ \\= \frac{\tan(180^\circ - B) + \tan B}{1 - \tan(180^\circ - B) \times \tan B} \\\ \\= \frac{-\tan B + \tan B}{1 + \tan B \times \tan B} \\\ \\= \frac{0}{1 + \tan^2 B} \\\ \\= 0$$​