Question

Prove the identity cos x/1 - sin x = sec x + tan x

Answer 1

Answer:

Proved

Step-by-step explanation:

Given

Prove that

$\frac{cos x}{1 - sin x} = sec x + tan x$

$\frac{cos x}{1 - sin x}$

Multiply the numerator and denominator by 1 + sinx

$\frac{cos x}{1 - sin x} * \frac{1 + sin x}{1 + sin x}$

Combine both fractions to form 1

$\frac{cos x (1 + sin x)}{(1 - sin x)(1 + sin x)}$

Expand the denominator using difference of two squares;

$i.e.\ (a - b)(a + b) = a^2 - b^2$

The expression becomes

$\frac{cos x (1 + sin x)}{(1^2 - sin^2 x)}$

$\frac{cos x (1 + sin x)}{(1 - sin^2 x)}$

From trigonometry; $1 - sin^2x = cos^2x$

The expression becomes

$\frac{cos x (1 + sin x)}{(cos^2 x)}$

Divide the numerator and the denominator by cos x

$\frac{(1 + sin x)}{(cos x)}$

Split fraction

$\frac{1}{cos x} + \frac{sin x}{cos x}$

From trigonometry; $\frac{1}{cos x} = sec x \ and\ \frac{sin\ x}{cos\ x} = tan\ x$

So;

$\frac{1}{cos x} + \frac{sin x}{cos x}$ = $sec x + tan x$