helpppppppp please.

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helpppppppp please.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Simplify the trigonometric expression. Show your work.
if you want to combine the two fractions, need to multiply by a common denominator
okay... idk how to do that?
\[[\frac{ 1 }{ 1+\sin \theta }+\frac{ 1 }{ 1-\sin \theta }]*\frac{ (1+\sin \theta)(1-\sin \theta) }{ (1+\sin \theta)(1-\sin \theta) }\]
Just multiplying the whole thing by 1... a ratio of something over itself
okay.. what does sin o stand for?
is it kinda like pi=3.14? does it stand for a number?
can u do it step by step and explain..
one sec
\(\bf \cfrac{1}{1+sin(\theta)}+\cfrac{1}{1-sin(\theta)}\impliedby LCD\to [1+sin(\theta)][1-sin(\theta)] \\ \quad \\ \cfrac{(1-sin(\theta))+(1+sin(\theta))}{[1+sin(\theta)][1-sin(\theta)]}\implies \cfrac{1\cancel{-sin(\theta)}+1\cancel{+sin(\theta)}}{[1+sin(\theta)][1-sin(\theta)]} \\ \quad \\ ----------------------------------\\ \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\qquad thus\\ ---------------------------------- \\ \quad \\ \cfrac{1\cancel{-sin(\theta)}+1\cancel{+sin(\theta)}}{{\color{brown}{ [1^2-sin^2(\theta)] }}}\implies \cfrac{2}{{\color{brown}{ cos^2(\theta)}}} \\ \quad \\ 2\cdot \cfrac{1}{cos^2(\theta)}\implies 2sec^2(\theta)\)
the denominator is combined, since it's just a difference of squares and keep in mind that \(\bf sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)={\color{brown}{ 1-cos^2(\theta)}}\)
woops, darn that came out off lemme redo that
\(\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)={\color{brown}{ 1-sin^2(\theta)}}\) rather

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