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Maths-

General

Easy

Question

If $tan invisible function application B equals fraction numerator n sin invisible function application A cos invisible function application A over denominator 1 minus n cos squared invisible function application A end fraction$ then tan (A + B) equals

$\frac{\sin A}{(1 - n) \cos A}$
$\frac{(n - 1) \cos A}{\sin A}$
$\frac{\sin A}{(n - 1) \cos A}$
$\frac{\sin A}{(n + 1) \cos A}$

The correct answer is: $fraction numerator sin invisible function application A over denominator left parenthesis 1 minus n right parenthesis cos invisible function application A end fraction$

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