Question

Please help me

Answer 1

By using algebra properties and trigonometric formulas we find that the trigonometric expression $\frac{1}{1 - \sin x} - \frac{1}{1 + \sin x}$ is equivalent to the trigonometric expression $\frac{2\cdot \tan x}{\cos x}$.

How to prove a trigonometric equivalence by algebraic and trigonometric procedures

In this question we have trigonometric expression whose equivalence to another expression has to be proved by using algebra properties and trigonometric formulas, including the fundamental trigonometric formula, that is, cos² x + sin² x = 1. Now we present in detail all steps to prove the equivalence:

$\frac{1}{1 - \sin x} - \frac{1}{1 + \sin x}$       Given.

$\frac{1 + \sin x - 1 + \sin x}{1 - \sin^{2}x}$      Subtraction between fractions with different denominator / (- 1) · a = - a.

$\frac{2\cdot \sin x}{\cos^{2}x}$      Definitions of addition and subtraction / Fundamental trigonometric formula (cos² x + sin² x = 1)

$\frac{2\cdot \tan x}{\cos x}$      Definition of tangent / Result

By using algebra properties and trigonometric formulas we conclude that the trigonometric expression $\frac{1}{1 - \sin x} - \frac{1}{1 + \sin x}$ is equal to the trigonometric expression $\frac{2\cdot \tan x}{\cos x}$. Hence, the former expression is equivalent to the latter one.

To learn more on trigonometric equations: brainly.com/question/10083069

#SPJ1