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anygoal [31]
3 years ago
11

How do you write cos, tan, and sec in terms of csc?

Mathematics
2 answers:
Katyanochek1 [597]3 years ago
8 0
\bf \textit{Pythagorean Identities}
\\ \quad \\
sin^2(\theta)+cos^2(\theta)=1
\\ \quad \\
1+cot^2(\theta)=csc^2(\theta)
\\ \quad \\
1+tan^2(\theta)=sec^2(\theta)\\\\
-----------------------------\\\\
csc(\theta)=\cfrac{1}{sin(\theta)}\qquad \qquad sec(\theta)=\cfrac{1}{cos(\theta)}

so hmm  let us use those ones

then

\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)
\\\\\\
cos(\theta)=\sqrt{1-sin^2(\theta)}\implies cos(\theta)=\sqrt{1-\frac{1}{csc^2(\theta)}}
\\\\\\
cos(\theta)=\sqrt{\cfrac{csc^2(\theta)-1}{csc^2(\theta)}}\\\\
-----------------------------\\\\

\bf 1+cot^2(\theta)=csc^2(\theta)\implies 1+\cfrac{1}{tan^2(\theta)}=csc^2(\theta)
\\\\\\
\cfrac{1}{tan^2(\theta)}=csc^2(\theta)-1\implies \cfrac{1}{csc^2(\theta)-1}=tan^2(\theta)
\\\\\\
\sqrt{\cfrac{1}{csc^2(\theta)-1}}=tan(\theta)\\\\
-----------------------------\\\\
sec(\theta)=\cfrac{1}{cos(\theta)}\implies sec(\theta)=\cfrac{1}{\sqrt{\frac{csc^2(\theta)-1}{csc^2(\theta)}}}
\\\\\\
sec(\theta)=\sqrt{\cfrac{csc^2(\theta)}{csc^2(\theta)-1}}
olga nikolaevna [1]3 years ago
4 0
Prolly not the best person to answer this question lol
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faust18 [17]

Answer:

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Step-by-step explanation:

The slope-intercept form of the equation of a line is

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As y intercept is -8, we have

y\:=\:mx\:-8

Since the line is parallel to 6x-y=7, then it has the same slope.

Let's put this equation in y\:=\:mx\:+\:b  form and find the slope

6x-y=7

y=6x-7

Thus, the slope is :

m = 6

Now we can finish the equation of the line in the slope-intercept form

y\:=\:mx\:+\:b

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Therefore, the equation of the line in the slope-intercept form is:

\:y=6x-8

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3 years ago
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Answer:

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