By using algebra properties and trigonometric formulas we find that the trigonometric expression [tex]\frac{1}{1 - \sin x} - \frac{1}{1 + \sin x}[/tex] is equivalent to the trigonometric expression [tex]\frac{2\cdot \tan x}{\cos x}[/tex].
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:
[tex]\frac{1}{1 - \sin x} - \frac{1}{1 + \sin x}[/tex] Given.
[tex]\frac{1 + \sin x - 1 + \sin x}{1 - \sin^{2}x}[/tex] Subtraction between fractions with different denominator / (- 1) · a = - a.
[tex]\frac{2\cdot \sin x}{\cos^{2}x}[/tex] Definitions of addition and subtraction / Fundamental trigonometric formula (cos² x + sin² x = 1)
[tex]\frac{2\cdot \tan x}{\cos x}[/tex] Definition of tangent / Result
By using algebra properties and trigonometric formulas we conclude that the trigonometric expression [tex]\frac{1}{1 - \sin x} - \frac{1}{1 + \sin x}[/tex] is equal to the trigonometric expression [tex]\frac{2\cdot \tan x}{\cos x}[/tex]. Hence, the former expression is equivalent to the latter one.
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