Answer :

Certainly! Let's simplify the given trigonometric expression step-by-step:

[tex]\[ \frac{\tan^2 A}{\tan A - 1} + \frac{\cot A}{1 - \tan A} \][/tex]

We start by recalling that [tex]\(\cot A\)[/tex] is the reciprocal of [tex]\(\tan A\)[/tex]:

[tex]\[ \cot A = \frac{1}{\tan A} \][/tex]

Now, substitute [tex]\(\cot A = \frac{1}{\tan A}\)[/tex] into the second term of the expression:

[tex]\[ \frac{\tan^2 A}{\tan A - 1} + \frac{\frac{1}{\tan A}}{1 - \tan A} \][/tex]

Simplify the second term by multiplying the numerator and the denominator by [tex]\(\tan A\)[/tex]:

[tex]\[ \frac{\tan^2 A}{\tan A - 1} + \frac{1}{\tan A (1 - \tan A)} \][/tex]

Next, notice that [tex]\(\tan A(1 - \tan A)\)[/tex] can be rearranged to [tex]\(\tan A\ - \tan^2 A\)[/tex]. Combining both fractions with common denominators would be one approach, but let's proceed differently.

To combine these fractions, we first observe the signs and simplifying trigonometric identities. We rewrite the expression in terms of a common trigonometric identity:

[tex]\[ \frac{\tan^2 A}{\tan A - 1} + \frac{1}{-\tan A (\tan A - 1)} \][/tex]

Factor out [tex]\((\tan A - 1)\)[/tex]:

[tex]\[ \frac{\tan^2 A}{\tan A - 1} - \frac{1}{\tan A (\tan A - 1)} \][/tex]

We now have a common denominator:

[tex]\[ \frac{\tan^3 A - 1}{\tan A (\tan A - 1)} \][/tex]

Simplify the numerator by factoring:

[tex]\[ \tan^3 A - 1 = (\tan A - 1)(\tan^2 A + \tan A + 1) \][/tex]

Thus,

[tex]\[ \frac{(\tan A - 1)(\tan^2 A + \tan A + 1)}{\tan A (\tan A - 1)} \][/tex]

When we cancel [tex]\(\tan A - 1\)[/tex]:

[tex]\[ \frac{\tan^2 A + \tan A + 1}{\tan A} \][/tex]

Separate the terms in the numerator:

[tex]\[ \frac{\tan^2 A}{\tan A} + \frac{\tan A}{\tan A} + \frac{1}{\tan A} \][/tex]

This simplifies each part separately:

[tex]\[ \tan A + 1 + \frac{1}{\tan A} \][/tex]

Recall that [tex]\(\frac{1}{\tan A} = \cot A\)[/tex]:

[tex]\[ \tan A + 1 + \cot A \][/tex]

Since [tex]\(\cot A = \frac{\cos A}{\sin A}\)[/tex] and [tex]\(\tan A = \frac{\sin A}{\cos A}\)[/tex], simplify terms:

[tex]\[ 1 + \frac{2}{\sin 2A} \][/tex]

So our final simplified expression is:

[tex]\[ 1 + \frac{2}{\sin 2A} \][/tex]