Animated Solution for Mathematics - Trigonometry: The expression $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$ can be written as

The beauty of trigonometric symmetry often lies in simplification. We are tasked with evaluating the expression:

At first glance, this appears to be a complex collection of ratios. However, the secret to solving it lies in the art of unification.

The first step in any complex trigonometric problem is to reduce the number of variables. We currently have both $\tan A$ and $\cot A$ present.

Since we know that $\cot A=\frac{1}{\tan A}$, our strategy is to convert everything into terms of $\tan A$. Substituting this identity into our expression yields:

Now, let us simplify the terms. For the first term, the denominator $1-\frac{1}{\tan A}$ simplifies to $\frac{\tan A-1}{\tan A}$.

When we invert the fraction, the first term becomes:

For the second term, the $\tan A$ in the numerator drops to the denominator, resulting in $\frac{1}{\tan A(1-\tan A)}$.

Many students stumble here. Notice that the denominators $(\tan A-1)$ and $(1-\tan A)$ are negatives of each other. By factoring out a $-1$ from the second term, we align them:

We now have a common denominator of $\tan A(\tan A-1)$.

Combining the fractions, the numerator becomes $\tan A\cdot {\tan }^{2}A-1$, which is ${\tan }^{3}A-1$. We are now looking at:

The numerator is a difference of cubes, $a^3-b^3$, where $a=\tan A$ and $b=1$. Using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$, we rewrite the numerator as:

The $(\tan A-1)$ term cancels out perfectly from the numerator and the denominator. We are left with:

By splitting the fraction, we obtain:

To reach the final form, we convert back to $\sin A$ and $\cos A$:

Since ${\sin }^{2}A+{\cos }^{2}A=1$, the expression simplifies to:

A seemingly chaotic expression has been reduced to a simple, elegant result.