Prove that $\left( \sec \theta -\cos \theta \right)\left( \cot \theta +\tan \theta \right)=\tan \theta \sec \theta $

Hint:Convert into sines and cosines and use $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $ and hence simplify the expression. Finally convert back to tangents and secants using $\dfrac{\sin \theta }{\cos \theta }=\tan \theta $ and $\dfrac{1}{\cos \theta }=\sec \theta $

Complete step-by-step answer:
Trigonometric ratios:
There are six trigonometric ratios defined on an angle of a right-angled triangle, viz sine, cosine, tangent, cotangent, secant and cosecant.
The sine of an angle is defined as the ratio of the opposite side to the hypotenuse.
The cosine of an angle is defined as the ratio of the adjacent side to the hypotenuse.
The tangent of an angle is defined as the ratio of the opposite side to the adjacent side.
The cotangent of an angle is defined as the ratio of the adjacent side to the opposite side.
The secant of an angle is defined as the ratio of the hypotenuse to the adjacent side.
The cosecant of an angle is defined as the ratio of the hypotenuse to the adjacent side.
Observe that sine and cosecant are multiplicative inverses of each other, cosine and secant are multiplicative inverses of each other, and tangent and cotangent are multiplicative inverses of each other.
Pythagorean identities:
The identities ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ and ${{\csc }^{2}}\theta =1+{{\cot }^{2}}\theta $ are known as Pythagorean identities as they are a direct consequence of Pythagoras’ theorem in a right angled triangle.
We have LHS $=\left( \sec \theta -\cos \theta \right)\left( \cot \theta +\tan \theta \right)$
We have $\sec \theta =\dfrac{1}{\sin \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta },\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.
Hence, we have
LHS $=\left( \dfrac{1}{\cos \theta }-\cos \theta \right)\left( \dfrac{\cos \theta }{\sin \theta }+\dfrac{\sin \theta }{\cos \theta } \right)$
Taking $\cos \theta $ as LCM in the first term and $\sin \theta \cos \theta $ as LCM in the second term, we have
LHS $=\dfrac{1-{{\cos }^{2}}\theta }{\cos \theta }\left( \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{\cos \theta \sin \theta } \right)$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $
Hence, we have
LHS $=\dfrac{{{\sin }^{2}}\theta }{\cos \theta }\left( \dfrac{1}{\sin \theta \cos \theta } \right)$
Hence, we have
LHS $=\dfrac{\sin \theta }{{{\cos }^{2}}\theta }$
Writing $\dfrac{\sin \theta }{{{\cos }^{2}}\theta }$ as $\dfrac{\sin \theta }{\cos \theta }\times \dfrac{1}{\cos \theta }$
Hence, we have
LHS $=\dfrac{\sin \theta }{\cos \theta }\times \dfrac{1}{\cos \theta }$
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$
Hence, we have
LHS $=\tan \theta \sec \theta $
Hence LHS = RHS.

Note: [1] In these types of questions dealing with sines and cosines is usually easier than tangents, cotangents, secants, cosecants etc. Hence we usually convert questions involving secants, tangents, cotangents and cosecants into sines and cosines and then simplify.
[2] Alternative Solution:
Converting $\cos\theta$ into $\sec\theta$ and $\cot\theta$ into $\tan\theta$
We have LHS $=\left( \sec \theta -\cos \theta \right)\left( \tan \theta +\cot \theta \right)=\left( \dfrac{{{\sec }^{2}}\theta -1}{\sec \theta } \right)\times \dfrac{1+{{\tan }^{2}}\theta }{\tan \theta }$
We know that ${{\sec }^{2}}\theta -1={{\tan }^{2}}\theta $ and $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $.
Hence, we have
LHS $=\dfrac{{{\tan }^{2}}\theta }{\sec \theta }\times \dfrac{{{\sec }^{2}}\theta }{\tan \theta }=\dfrac{{{\left( \sec \theta \tan \theta \right)}^{2}}}{\sec \theta \tan \theta }=\sec \theta \tan \theta $
Hence LHS = RHS