How do you verify the identity \[\cot \alpha + \tan \alpha = \cos ec\alpha \sec \alpha \]

Hint: This problem deals with solving the given trigonometric equation with the help of the basic trigonometric identities such as given below:
$ {\cos ^2}\alpha + {\sin ^2}\alpha = 1$
The reciprocal of the trigonometric identities such as shown below:
$ \Rightarrow \tan \alpha = \dfrac{1}{{\cot \alpha }}$
$ \Rightarrow \sin \alpha = \dfrac{1}{{\cos ec \alpha }}$
$ \Rightarrow \cos \alpha = \dfrac{1}{{\sec \alpha }}$

Complete step by step solution:
Given the left hand side of the equation is \[\cot \alpha + \tan \alpha \]
Consider the equation \[\cot \alpha + \tan \alpha \] as shown below:
$ \Rightarrow \cot \alpha + \tan \alpha $
We know that the reciprocal of $\cot \alpha $ is equal to $\tan \alpha $, which is given by: $\cot \alpha = \dfrac{1}{{\tan \alpha }}$.
$ \Rightarrow \cot \alpha + \tan \alpha = \dfrac{1}{{\tan \alpha }} + \tan \alpha $
Simplifying the right hand side of the above equation as shown below:
$ \Rightarrow \cot \alpha + \tan \alpha = \dfrac{{1 + {{\tan }^2}\alpha }}{{\tan \alpha }}$
Now converting the ratio $\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}$ as shown below:
$ \Rightarrow \cot \alpha + \tan \alpha = \dfrac{{1 + \dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}}{{\dfrac{{\sin \alpha }}{{\cos \alpha }}}}$
On further simplification of the right hand side of the above equation as shown below:
$ \Rightarrow \cot \alpha + \tan \alpha = \dfrac{{\dfrac{{{{\cos }^2}\alpha + {{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}}{{\dfrac{{\sin \alpha }}{{\cos \alpha }}}}$
$ \Rightarrow \cot \alpha + \tan \alpha = \dfrac{{{{\cos }^2}\alpha + {{\sin }^2}\alpha }}{{\sin \alpha \cos \alpha }}$
We know the basic trigonometric identity which is the sum of squares of cosine angle and the sine angle is always equal to unity, which is given by: ${\cos ^2}\alpha + {\sin ^2}\alpha = 1$, substituting it as shown below:
$ \Rightarrow \cot \alpha + \tan \alpha = \dfrac{1}{{\sin \alpha \cos \alpha }}$
We know that the reciprocal of $\sin \alpha $ is equal to $\cos ec\alpha $, which is given by: $\dfrac{1}{{\sin \alpha }} = \cos ec\alpha $ and whereas the reciprocal of $\cos \alpha $ is equal to $\sec \alpha $, which is given by: $\dfrac{1}{{\cos \alpha }} = \sec \alpha $, now substituting these identities in the above equation as shown below:
\[ \Rightarrow \cot \alpha + \tan \alpha = \cos ec\alpha \sec \alpha \]
Hence proved.

Note: Please note that while solving this problem some basic trigonometric identities are used such as the sum of the squares of cosine angle and the sine angle is always equal to unity, which is given by: ${\cos ^2}\alpha + {\sin ^2}\alpha = 1$. Similarly there are other basic trigonometric identities such as the difference of the squares of secant angle and the tangent angle is equal to unity, similarly for cosecant and cotangent angle as shown below:
$ \Rightarrow {\sec ^2}\alpha - {\tan ^2}\alpha = 1$
$ \Rightarrow \cos e{c^2}\alpha - {\cot ^2}\alpha = 1$
The reciprocal identities are given by:
$ \Rightarrow \cos ec\alpha = \dfrac{1}{{\sin \alpha }}$
$ \Rightarrow \sec \alpha = \dfrac{1}{{\cos \alpha }}$
$ \Rightarrow \cot \alpha = \dfrac{1}{{\tan \alpha }}$