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Ierofanga [76]
3 years ago
10

List the elements in the set . Let U = {q, r, s, t, u, v, w, x, y, z} A = {q, s, u, w, y} B = {q, s, y, z} C = {v, w, x, y, z}.

(B ∪ A)'
Mathematics
1 answer:
zubka84 [21]3 years ago
4 0
I don’t get the way you ordered this can you sent me a picture of it or some
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By using <em>algebra</em> properties and <em>trigonometric</em> formulas we find that the <em>trigonometric</em> expression \frac{1}{1 - \sin x} - \frac{1}{1 + \sin x} is equivalent to the <em>trigonometric</em> expression \frac{2\cdot \tan x}{\cos x}.

<h3>How to prove a trigonometric equivalence by algebraic and trigonometric procedures</h3>

In this question we have <em>trigonometric</em> expression whose equivalence to another expression has to be proved by using <em>algebra</em> properties and <em>trigonometric</em> formulas, including the <em>fundamental trigonometric</em> 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 <em>algebra</em> properties and <em>trigonometric</em> formulas we conclude that the <em>trigonometric</em> expression \frac{1}{1 - \sin x} - \frac{1}{1 + \sin x} is equal to the <em>trigonometric</em> 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

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