anonymous
  • anonymous
helpppppppp please.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
@Hero
DanJS
  • DanJS
What do you want to do with that?

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anonymous
  • anonymous
Simplify the trigonometric expression. Show your work.
DanJS
  • DanJS
if you want to combine the two fractions, need to multiply by a common denominator
anonymous
  • anonymous
okay... idk how to do that?
DanJS
  • DanJS
\[[\frac{ 1 }{ 1+\sin \theta }+\frac{ 1 }{ 1-\sin \theta }]*\frac{ (1+\sin \theta)(1-\sin \theta) }{ (1+\sin \theta)(1-\sin \theta) }\]
DanJS
  • DanJS
Just multiplying the whole thing by 1... a ratio of something over itself
anonymous
  • anonymous
okay.. what does sin o stand for?
anonymous
  • anonymous
is it kinda like pi=3.14? does it stand for a number?
anonymous
  • anonymous
can u do it step by step and explain..
jdoe0001
  • jdoe0001
one sec
jdoe0001
  • jdoe0001
\(\bf \cfrac{1}{1+sin(\theta)}+\cfrac{1}{1-sin(\theta)}\impliedby LCD\to [1+sin(\theta)][1-sin(\theta)] \\ \quad \\ \cfrac{(1-sin(\theta))+(1+sin(\theta))}{[1+sin(\theta)][1-sin(\theta)]}\implies \cfrac{1\cancel{-sin(\theta)}+1\cancel{+sin(\theta)}}{[1+sin(\theta)][1-sin(\theta)]} \\ \quad \\ ----------------------------------\\ \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\qquad thus\\ ---------------------------------- \\ \quad \\ \cfrac{1\cancel{-sin(\theta)}+1\cancel{+sin(\theta)}}{{\color{brown}{ [1^2-sin^2(\theta)] }}}\implies \cfrac{2}{{\color{brown}{ cos^2(\theta)}}} \\ \quad \\ 2\cdot \cfrac{1}{cos^2(\theta)}\implies 2sec^2(\theta)\)
jdoe0001
  • jdoe0001
the denominator is combined, since it's just a difference of squares and keep in mind that \(\bf sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)={\color{brown}{ 1-cos^2(\theta)}}\)
jdoe0001
  • jdoe0001
woops, darn that came out off lemme redo that
jdoe0001
  • jdoe0001
\(\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)={\color{brown}{ 1-sin^2(\theta)}}\) rather

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