How do you prove $ \sec x\left( {{\csc }^{2}}x \right)-{{\csc }^{2}}x=\dfrac{\sec x}{1+\cos x} $ ?

Hint: We first take common out from the left-hand part of the equation of $ \sec x\left( {{\csc }^{2}}x \right)-{{\csc }^{2}}x $ . Then we simplify the equation. We convert the denominator using the relation $ \sec x=\dfrac{1}{\cos x} $ . Then we use the theorem \[{{\cos }^{2}}x+{{\sin }^{2}}x=1\]. We eliminate the part $ \left( 1-\cos x \right) $ . After elimination we get the right-hand side of the equation.

Complete step by step answer:
We have to prove the trigonometric equation $ \sec x\left( {{\csc }^{2}}x \right)-{{\csc }^{2}}x=\dfrac{\sec x}{1+\cos x} $ .
We take the left-hand side of the equation $ \sec x\left( \cos e{{c}^{2}}x \right)-\cos e{{c}^{2}}x $ and prove the right-side part. We can take $ {{\csc }^{2}}x $ as common from both terms.
We get $ \sec x\left( {{\csc }^{2}}x \right)-{{\csc }^{2}}x={{\csc }^{2}}x\left( \sec x-1 \right) $ .
We solve the equation inside the brackets.
The part is $ \left( \sec x-1 \right) $ . We know that $ \sec x=\dfrac{1}{\cos x} $ .
Therefore, $ \left( \sec x-1 \right)=\dfrac{1}{\cos x}-1=\dfrac{1-\cos x}{\cos x} $ .
We know the relation between $ \csc x $ and $ \sin x $ is $ \csc x=\dfrac{1}{\sin x} $ .
This relation gives $ {{\csc }^{2}}x=\dfrac{1}{{{\sin }^{2}}x} $ .
The revised form of the equation is $ {{\csc }^{2}}x\left( \sec x-1 \right)=\dfrac{1}{{{\sin }^{2}}x}\left( \dfrac{1-\cos x}{\cos x} \right) $ .
We now use the identity theorem of trigonometry $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ which gives us $ {{\sin }^{2}}x=1-{{\cos }^{2}}x $ . We place the value in the equation and get
 $ \dfrac{1}{{{\sin }^{2}}x}\left( \dfrac{1-\cos x}{\cos x} \right)=\dfrac{1}{1-{{\cos }^{2}}x}\times \dfrac{\left( 1-\cos x \right)}{\cos x} $ .
The denominator is in the form of $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ . We now apply the identity theorem for the term $ 1-{{\cos }^{2}}x $ . We assume the values $ a=1;b=\cos x $ .
Applying the theorem, we get $ 1-{{\cos }^{2}}x=\left( 1+\cos x \right)\left( 1-\cos x \right) $ .
The equation becomes $ \dfrac{1}{1-{{\cos }^{2}}x}\times \dfrac{\left( 1-\cos x \right)}{\cos x}=\dfrac{1}{\left( 1-\cos x \right)\left( 1+\cos x \right)}\times \dfrac{\left( 1-\cos x \right)}{\cos x} $ .
We can now eliminate the $ \left( 1-\cos x \right) $ from both denominator and numerator.
The equation becomes $ \dfrac{1}{\left( 1-\cos x \right)\left( 1+\cos x \right)}\times \dfrac{\left( 1-\cos x \right)}{\cos x}=\dfrac{1}{\cos x\left( 1+\cos x \right)} $ .
Now we apply the theorem $ \sec x=\dfrac{1}{\cos x} $ again to convert the $ \cos x $ .
The final form is $ \dfrac{1}{\cos x\left( 1+\cos x \right)}=\dfrac{\sec x}{\left( 1+\cos x \right)} $ .
Thus proved $ \sec x\left( {{\csc }^{2}}x \right)-{{\csc }^{2}}x=\dfrac{\sec x}{1+\cos x} $.

Note:
 It is important to remember that the condition to eliminate the $ \left( 1-\cos x \right) $ from both denominator and numerator is $ \left( 1-\cos x \right)\ne 0 $ . No domain is given for the variable $ x $ . The simplified condition will be $ x\ne n\pi \pm \dfrac{\pi }{2},n\in \mathbb{Z} $ .