Prove that $\left( \csc \phi -\sin \phi \right)\left( \sec \phi -\cos \phi \right)=\dfrac{1}{\tan \phi +\cot \phi }$.

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

Complete step by step answer:
We try to simplify the trigonometric equation \[\left( \csc \phi -\sin \phi \right)\].
We get \[\csc \phi -\sin \phi \]. We know that $\csc x=\dfrac{1}{\sin x}$. We also have $1-{{\sin }^{2}}x={{\cos }^{2}}x$.
Therefore, $\csc \phi -\sin \phi =\dfrac{1}{\sin \phi }-\sin \phi =\dfrac{1-{{\sin }^{2}}\phi }{\sin \phi }=\dfrac{{{\cos }^{2}}\phi }{\sin \phi }$.
We have to prove the trigonometric equation $\sec \phi -\cos \phi $.
We know that $\sec \phi =\dfrac{1}{\cos \phi }$. We also have $1-{{\cos }^{2}}x={{\sin }^{2}}x$.
Therefore, $\sec \phi -\cos \phi =\dfrac{1}{\cos \phi }-\cos \phi =\dfrac{1-{{\cos }^{2}}\phi }{\cos \phi }=\dfrac{{{\sin }^{2}}\phi }{\cos \phi }$.

The multiplication gives
$\left( \csc \phi -\sin \phi \right)\left( \sec \phi -\cos \phi \right) =\dfrac{{{\cos }^{2}}\phi }{\sin \phi }\times \dfrac{{{\sin }^{2}}\phi }{\cos \phi } \\
\Rightarrow \left( \csc \phi -\sin \phi \right)\left( \sec \phi -\cos \phi \right)=\sin \phi \cos \phi \\ $
We write the simplification as $\sin \phi \cos \phi =\dfrac{\sin \phi \cos \phi }{1}=\dfrac{\sin \phi \cos \phi }{{{\sin }^{2}}\phi +{{\cos }^{2}}\phi }$.
Now we rearrange the expression as
$\dfrac{\sin \phi \cos \phi }{{{\sin }^{2}}\phi +{{\cos }^{2}}\phi }=\dfrac{1}{\dfrac{{{\sin }^{2}}\phi +{{\cos }^{2}}\phi }{\sin \phi \cos \phi }}$
Completing the division, awe get
\[\dfrac{1}{\dfrac{{{\sin }^{2}}\phi +{{\cos }^{2}}\phi }{\sin \phi \cos \phi }}=\dfrac{1}{\dfrac{\sin \phi }{\cos \phi }+\dfrac{\cos \phi }{\sin \phi }} \\
\therefore \dfrac{1}{\dfrac{{{\sin }^{2}}\phi +{{\cos }^{2}}\phi }{\sin \phi \cos \phi }}=\dfrac{1}{\tan \phi +\cot \phi }\]

Thus proved $\left( \csc \phi -\sin \phi \right)\left( \sec \phi -\cos \phi \right)=\dfrac{1}{\tan \phi +\cot \phi }$.

Note: It is important to remember that the condition to eliminate the $\sin x$ from both denominator and numerator is $\sin x\ne 0$. No domain is given for the variable $x$. The simplified condition will be $x\ne n\pi ,n\in \mathbb{Z}$. The identities ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and $\sec x=\dfrac{1}{\cos x}$ are valid for any value of $x$. The division of the fraction part only gives $\sin x$ as the solution.