We will simplify the left hand side of the equation to look like the right
[tex]\displaystyle \frac{1-sin(t)}{cos^{2}(t)} + \frac{1}{1 - sin(t)}[/tex]
taking the LCM
[tex]\displaystyle \frac{[1-sin(t)]^{2} + cos^{2}(t)}{[cos^{2}(t)][1 - sin(t)]}[/tex]
expanding the binomial in the numerator
[tex]\displaystyle \frac{[1 + sin^{2}(t) - 2sin(t) + cos^{2}(t)]}{[cos^{2}(t)][1 - sin(t)]}[/tex]
Since sin²x + cos²x = 1
[tex]\displaystyle \frac{[2 - 2sin(t)]}{[cos^{2}(t)][1 - sin(t)]}[/tex]
factoring out the 2 from the numerator
[tex]\displaystyle \frac{2[1 - sin(t)]}{[cos^{2}(t)][1 - sin(t)]}[/tex]
1-sin(t) will cancel out
[tex]\displaystyle \frac{2}{cos^{2}(t)}[/tex]
since cos²(t) = 1/(sec²(t))
[tex]2 sec^{2}(t)[/tex]
which is equal to the right hand side of our given equation.
So we verified the identity!
