Answer:
The Pythagorean identity states that
[tex]\sin^2 t + \cos^2 t = 1[/tex]
Using that we can rewrite the left denominator as:
[tex]1 - \sin^2 t[/tex]
Which can be factored as
[tex](1 - \sin t)(1+ \sin t)[/tex]
The numerator we can expand as:
[tex](1 - \sin t)(1 - \sin t)[/tex]
On the right hand side, let's multply numerator and denominator with (1 - sin t):
The total formula then becomes:
[tex]\dfrac{(1-\sin t)(1-\sin t)}{(1 - \sin t)(1 + \sin t)} = \dfrac{(1-\sin t)(1 - \sin t)}{(1+\sin t)(1 - \sin t)}[/tex]
There you go... left and right are equal.
