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

Hint: First we will evaluate the right-hand of the equation and then further the left-hand side of the equation. We will use the following formula $ \cot x = \dfrac{{\cos x}}{{\sin x}}\,\, $ and $ \csc x = \dfrac{1}{{\sin x}} $
to evaluate and then we will further simplify this expression form and hence evaluate the value of the term.

Complete step-by-step answer:
We will start off by using the formula
 $ \cot x = \dfrac{{\cos x}}{{\sin x}}\,\, $ and $ \csc x = \dfrac{1}{{\sin x}} $ .
Here, we will start by evaluating the right-hand side of the equation.
Hence, the equation will become,
 $
   = {\cot ^2}x - {\csc ^2}x \\
   = \dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}} - \dfrac{1}{{{{\sin }^2}x}} \\
   = \dfrac{{{{\cos }^2}x - 1}}{{{{\sin }^2}x}} \;
  $
Now we know the identity $ {\sin ^2}x + {\cos ^2}x = 1 $ which can also be written as $ {\sin ^2}x = 1 - {\cos ^2}x $ .
Therefore, the expression will become,
 $
   = \dfrac{{ - {{\sin }^2}x}}{{{{\sin }^2}x}} \\
   = - 1 \;
  $
Hence, the value of the expression $ {\cot ^2}x - {\csc ^2}x $ is $ - 1 $ .
So, the correct answer is “-1”.

Note: While choosing the side to solve, always choose the side where you can directly apply the trigonometric identities. Also, remember the trigonometric identities $ {\sin ^2}x + {\cos ^2}x = 1 $ and $ \cos 2x = 2{\cos ^2}x - 1 $ . While opening the brackets make sure you are opening the brackets properly with their respective signs. Also remember that $ \tan x = \,\dfrac{{\sin x}}{{\cos x}} $ .
While applying the double angle identities, first choose the identity according to the terms you have then choose the terms from the expression involving which you are using the double angle identities. While modifying any identity make sure that when you back trace the identity, you get the same original identity.