How do you simplify ${{\tan }^{2}}x-{{\cot }^{2}}x$?

Hint: In this problem we need to simplify the given trigonometric equation. For this we will convert the whole equation in terms of $\sin x$, $\cos x$ by using the basic definitions of trigonometric ratios i.e., $\tan x=\dfrac{\sin x}{\cos x}$, $\cot x=\dfrac{\cos x}{\sin x}$. Now we will substitute those values in the given equation and simplify them by taking the LCM and calculating the difference. Now we will apply the algebraic formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ and use the trigonometric identities and formulas to simplify the obtained equation. After simplifying the equation, we will get our required solution.

Complete step by step answer:
Given that, ${{\tan }^{2}}x-{{\cot }^{2}}x$.
From the basic definitions of trigonometric ratios substituting the values $\tan x=\dfrac{\sin x}{\cos x}$, $\cot x=\dfrac{\cos x}{\sin x}$ in the above equation, then we will get
$\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x={{\left( \dfrac{\sin x}{\cos x} \right)}^{2}}-{{\left( \dfrac{\cos x}{\sin x} \right)}^{2}}$
Simplifying the above equation, then we will get
$\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}-\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}$
Taking the LCM and doing the subtraction in the above equation, then we will get
$\begin{align}
  & \Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=\dfrac{{{\sin }^{2}}x\times {{\sin }^{2}}x-{{\cos }^{2}}x\times {{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x} \\
 & \Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=\dfrac{{{\left( {{\sin }^{2}}x \right)}^{2}}-{{\left( {{\cos }^{2}}x \right)}^{2}}}{{{\left( \sin x\cos x \right)}^{2}}} \\
\end{align}$
Using the algebraic formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the above equation, then we will get
$\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=\dfrac{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)}{{{\left( \sin x\cos x \right)}^{2}}}$
We have the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, trigonometric formulas ${{\sin }^{2}}x-{{\cos }^{2}}x=-\cos 2x$, $\sin x\cos x=\dfrac{\sin 2x}{2}$. Substituting these values in the above equation, then we will get
$\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=\dfrac{\left( 1 \right)\left( -\cos 2x \right)}{{{\left( \dfrac{\sin 2x}{2} \right)}^{2}}}$
Simplifying the above equation, then we will get
$\begin{align}
  & \Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=-\dfrac{\cos 2x}{\dfrac{{{\sin }^{2}}2x}{4}} \\
 & \Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=-4\dfrac{\cos 2x}{{{\sin }^{2}}2x} \\
\end{align}$
Writing the denominator ${{\sin }^{2}}2x=\sin 2x.\sin 2x$, then we will get
$\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=-4\times \dfrac{\cos 2x}{\sin 2x}\times \dfrac{1}{\sin 2x}$
We have the trigonometric formulas $\dfrac{\cos x}{\sin x}=\cot x$, $\dfrac{1}{\sin x}=\csc x$. Applying those formulas in the above equation, then we will get
$\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=-4\cot 2x\csc 2x$

Hence the simplified form of the given equation ${{\tan }^{2}}x-{{\cot }^{2}}x$ is $-4\cot 2x\csc 2x$.

Note: In some cases, students may use the trigonometric identities ${{\sec }^{2}}x-{{\tan }^{2}}x=1$, ${{\csc }^{2}}x-{{\cot }^{2}}x=1$ to simplify the equation. From the identity ${{\sec }^{2}}x-{{\tan }^{2}}x=1$, the value of ${{\tan }^{2}}x$ is ${{\sec }^{2}}x-1$ and from the identity ${{\csc }^{2}}x-{{\cot }^{2}}x=1$, the value of ${{\cot }^{2}}x$ is ${{\csc }^{2}}x-1$. Substituting these values in the above equation, then we will get
$\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=\left( {{\sec }^{2}}x-1 \right)-\left( {{\csc }^{2}}x-1 \right)$
Simplifying the above equation, then we will get
$\begin{align}
  & \Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x={{\sec }^{2}}x-1-{{\csc }^{2}}x+1 \\
 & \Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x={{\sec }^{2}}x-{{\csc }^{2}}x \\
\end{align}$
I think the above form is not the simplified form of the given equation. So, I suggest students don’t follow the procedure in this note. Please use the procedure which we discussed in the solution part.