How do you simplify the expression ${\csc ^2}x$ - 1?

Hint: Use the formula of Pythagorean identity: \[{\sin ^2}x\]+ ${\cos ^2}x$ = 1,

Complete step by step solution:
As we know according to Pythagorean identity: \[{\sin ^2}x\]+ ${\cos ^2}x$ = 1
Now, let us divide both the side by \[{\sin ^2}x\]
$\Rightarrow$ $\dfrac{{{{\sin }^2}x}}{{{{\sin }^2}x}}$ + \[\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}\]= 1 ……….equation (1)
Now, as we know by trigonometric quotient identity $\dfrac{{\cos x}}{{\sin x}}$= $\cot x$
Squaring both the sides
 $\Rightarrow$ $\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}}$ = ${\cot ^2}x$……….equation (2)
And also we know by reciprocal identity $\dfrac{1}{{\sin x}}$ = $\csc x$
Now, squaring both the sides –
$\dfrac{1}{{{{\sin }^2}x}}$ = ${\csc ^2}x$……….equation (3)
Now, putting the values of equation (2) and equation (3) in equation (1)
$\Rightarrow$ $\dfrac{{{{\sin }^2}x}}{{{{\sin }^2}x}}$ + ${\cot ^2}x$= ${\csc ^2}x$
$\Rightarrow$ 1 + ${\cot ^2}x$= ${\csc ^2}x$
Now, rearranging the above equation:
$\Rightarrow$ ${\cot ^2}x$ = ${\csc ^2}x$- 1

Therefore, the solution of the expression ${\csc ^2}x$- 1 = ${\cot ^2}x$

Note:
The quotient identities define the relationship among certain trigonometric functions and can be very helpful in verifying other identities:
\[\cot x\] = $\dfrac{{\cos x}}{{\sin x}}$
$\tan x$= $\dfrac{{\sin x}}{{\cos x}}$