The ratio of the length to the width of

Mathematics

1 answer:

tekilochka [14]3 years ago

8 0

What are you trying to find out the length of

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How do I get the answer

EleoNora [17]

$\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\\\
sec(\theta)=\cfrac{1}{cos(\theta)}\\\\
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\cfrac{1}{1+sin(\theta)}+\cfrac{1}{1-sin(\theta)}\impliedby LCD=[1-sin(\theta)][1+sin(\theta)]
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\cfrac{1\underline{-sin(\theta)}+1\underline{+sin(\theta)}}{[1-sin(\theta)][1+sin(\theta)]}\implies 
\cfrac{2}{[1-sin(\theta)][1+sin(\theta)]}\\\\
-------------------------------\\\\$

$\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\
\cfrac{2}{1^2-sin^2(\theta)}\implies \cfrac{2}{1-sin^2(\theta)}\implies \cfrac{2}{cos^2(\theta)}
\\\\\\
\cfrac{2}{1}\cdot \cfrac{1^2}{cos^2(\theta)}\implies 2sec^2(\theta)$

5 0

3 years ago

Read 2 more answers

Please need help thank

babymother [125]

Answer:

Step-by-step explanation:

One member in the graph only went one time which is fewer than two times.

3 0

3 years ago

Ron is making a scale drawing to enlarge a picture. The original dimensions of the picture are 8 inches wide by 64 inches long.

Levart [38]