Answer:
See below.
Step-by-step explanation:
I'm going to use what I assume is the correct question.
[tex] \dfrac{1 - \sin x}{\cos x} = \dfrac{\cos x}{1 + \sin x} [/tex]
[tex] \dfrac{\cos x}{\cos x} \times \dfrac{1 - \sin x}{\cos x} = \dfrac{\cos x}{1 + \sin x} [/tex]
[tex] \dfrac{\cos x(1 - \sin x)}{\cos^2 x} = \dfrac{\cos x}{1 + \sin x} [/tex]
Now use the identity
[tex] sin^2 x + cos^2 x = 1 [/tex]
[tex] cos^2 x = 1 - sin^2 x [/tex]
We replace [tex] \cos^2 x [/tex] in the left denominator.
[tex] \dfrac{\cos x(1 - \sin x)}{1 - \sin^2 x} = \dfrac{\cos x}{1 + \sin x} [/tex]
Factor the difference of squares in the left side denominator.
[tex] \dfrac{\cos x(1 - \sin x)}{(1 + \sin x)(1 - \sin x)} = \dfrac{\cos x}{1 + \sin x} [/tex]
[tex] \dfrac{\cos x}{1 + \sin x} = \dfrac{\cos x}{1 + \sin x} [/tex]
