Prove that
\[\cos ecx\left( \sec x-1 \right)-\cot x\left( 1-\cos x \right)=\tan x-\sin x\]

Hint: In this type of question, we can first convert either the left hand side that is L.H.S or the right hand side that is R.H.S into the most basic form that is in terms of sin and cos functions.
For this, we can use the trigonometric relations to convert the cosec, sec, tan and cot functions in terms of sin and cos functions which can be done using the following relations
\[\begin{align}
  & \tan x=\dfrac{\sin x}{\cos x} \\
 & \cot x=\dfrac{\cos x}{\sin x} \\
 & \cos ecx=\dfrac{1}{\sin x} \\
 & \sec x=\dfrac{1}{\cos x} \\
\end{align}\]

Complete step-by-step answer:
As mentioned in the question, we have to prove the given statement.
Now, as mentioned in the hint, we have to first convert every involved trigonometric function into cos and sin functions and then we can proceed further.
Now, on using the relations given in the hint, we can write the L.H.S as follows
\[\begin{align}
  & =\cos ecx\left( \sec x-1 \right)-\cot x\left( 1-\cos x \right) \\
 & =\dfrac{1}{\sin x}\left( \dfrac{1}{\cos x}-1 \right)-\dfrac{\cos x}{\sin x}\left( 1-\cos x \right) \\
 & =\dfrac{\left( 1-\cos x \right)}{\sin x\cdot \cos x}-\dfrac{\left( \cos x-{{\cos }^{2}}x \right)}{\sin x} \\
 & =\dfrac{\left( 1-\cos x \right)}{\sin x\cdot \cos x}-\dfrac{\left( \cos x-{{\cos }^{2}}x \right)\cos x}{\sin x\cdot \cos x} \\
 & =\dfrac{\left( 1-\cos x \right)}{\sin x\cdot \cos x}\left( 1-{{\cos }^{2}}x \right) \\
 & =\dfrac{\left( 1-\cos x \right){{\sin }^{2}}x}{\sin x\cdot \cos x} \\
 & =\dfrac{\left( 1-\cos x \right)\sin x}{\cos x} \\
 & =\dfrac{\sin x}{\cos x}-\sin x \\
 & =\tan x-\sin x \\
\end{align}\]
Hence, as L.H.S is equal to R.H.S, therefore, the statement is proved.

Note: The students can convert the R.H.S into the L.H.S and then prove the statement. It could also be done using the above relations.