Answer:
Proved
Step-by-step explanation:
Given
Prove that
[tex]\frac{cos x}{1 - sin x} = sec x + tan x[/tex]
[tex]\frac{cos x}{1 - sin x}[/tex]
Multiply the numerator and denominator by 1 + sinx
[tex]\frac{cos x}{1 - sin x} * \frac{1 + sin x}{1 + sin x}[/tex]
Combine both fractions to form 1
[tex]\frac{cos x (1 + sin x)}{(1 - sin x)(1 + sin x)}[/tex]
Expand the denominator using difference of two squares;
[tex]i.e.\ (a - b)(a + b) = a^2 - b^2[/tex]
The expression becomes
[tex]\frac{cos x (1 + sin x)}{(1^2 - sin^2 x)}[/tex]
[tex]\frac{cos x (1 + sin x)}{(1 - sin^2 x)}[/tex]
From trigonometry; [tex]1 - sin^2x = cos^2x[/tex]
The expression becomes
[tex]\frac{cos x (1 + sin x)}{(cos^2 x)}[/tex]
Divide the numerator and the denominator by cos x
[tex]\frac{(1 + sin x)}{(cos x)}[/tex]
Split fraction
[tex]\frac{1}{cos x} + \frac{sin x}{cos x}[/tex]
From trigonometry; [tex]\frac{1}{cos x} = sec x \ and\ \frac{sin\ x}{cos\ x} = tan\ x[/tex]
So;
[tex]\frac{1}{cos x} + \frac{sin x}{cos x}[/tex] = [tex]sec x + tan x[/tex]
