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Lunna [17]
3 years ago
5

The local meteorologist is converting temperatures from Celsius to Fahrenheit. she knows that the formula for the conversion fro

m Fahrenheit to Celsius is C=5/9(f-32) what would be the formula for converting to Fahrenheit from Celsius?
Mathematics
1 answer:
andreev551 [17]3 years ago
6 0
In this equation, we are solving for f. The first thing we have to do is get rid of the denominator, so we have to multiply both sides by 9, which leaves us with 9C=5*(f-32). Then we divide both sides by 5, which is 9/5C=f-32. The last thing we have to do is add 32 to both sides to isolate f, which gives us our final equation of (5/9)C+32=f. 
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The <em>trigonometric</em> expression \frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1} is equivalent to the <em>trigonometric</em> expression \sec \alpha \cdot \csc \alpha + 1.

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that <em>one</em> side of the equality is equal to the expression of the <em>other</em> side, requiring the use of <em>algebraic</em> and <em>trigonometric</em> properties. Now we proceed to present the corresponding procedure:

\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}

\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }

\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}

\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}

\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}

\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}

\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}

\tan \alpha + 1 + \cot \alpha

\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1

\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1

\frac{1}{\cos \alpha \cdot \sin \alpha} + 1

\sec \alpha \cdot \csc \alpha + 1

The <em>trigonometric</em> expression \frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1} is equivalent to the <em>trigonometric</em> expression \sec \alpha \cdot \csc \alpha + 1.

To learn more on trigonometric expressions: brainly.com/question/10083069

#SPJ1

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2 years ago
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