Easy Trigonometry

$(\dfrac{1}{Cos a}-\dfrac{Sin a}{Cos a}) × (\dfrac{1}{Cos a}-\dfrac{Sin a}{Cos a})$
$(\dfrac{1-Sin a}{Cos a}) × (\dfrac{1+Sin a}{Cos a})$
$\dfrac{1-Sin a^2}{Cos a^2}$
$\dfrac{Cos a^2}{Cos a^2}=\boxed{1}$ [As $Cos a^2+ Sin a^2=1$ ]

It is entirely possible to solve this without knowing any trigonometric identities.

Create a right-angle triangle that has three sides x, y, z (y being the hypotenuse) and an angle a, such that:

$\sin { a } = \frac { x }{ y }$ $\cos { a } = \frac { z }{ y }$ $\tan { a } = \frac { x }{ z }$

Substitute their equivalents into the problem,

$(\frac { y }{ z } -\frac { \frac { x }{ y } }{ \frac { z }{ y } } )(\frac { y }{ z } +\frac { \frac { x }{ y } }{ \frac { z }{ y } } ) = (\frac { y }{ z } -\frac { x }{ z } )(\frac { y }{ z } +\frac { x }{ z } ) = \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { z }^{ 2 } }$

This is a right-angle triangle, so the Pythagoras theorem applies, therefore,

${ x }^{ 2 } + { z }^{ 2 } = { y }^{ 2 }$ ${ z }^{ 2 } = { y }^{ 2 } - { x }^{ 2 }$

Thus,

$\frac { { y }^{ 2 }-{ x }^{ 2 } }{ { z }^{ 2 } } = \frac { { z }^{ 2 } }{ { z }^{ 2 } } = \boxed{1}$

$\begin{aligned} &= \bigg(\frac{1}{\cos a} - \frac{\sin a}{\cos a}\bigg) \bigg(\frac{1}{\cos a} + \frac{\sin a}{\cos a}\bigg) \\ &= (\sec a - \tan a)(\sec a + \tan a) \\ &= \sec^2 a - \tan^2 a \\ &= \boxed{1} \end{aligned}$