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Show that $\frac{\csc x}{\cos x} - \frac{\cos x}{sin x} = \tan x$ is an identity.

First find a common denominator and change each fraction so it has that common denominator: $\frac{\csc x}{\cos x} - \frac{\cos x}{\sin x} = \tan x$

$\frac{\csc x}{\cos x} \cdot \frac{\sin x}{\sin x} - \frac{\cos x}{\sin x} \cdot \frac{\cos x}{\cos x} = \frac{\csc x \sin x}{\cos x \sin x} - \frac{\cos^{2} x}{\cos x \sin x} =$

"Simplfy the numerator in the first fraction. Subtract the two fractions, writing the two terms in the same numerator. Then replace the numerator using the appropriate form of the Pythagorean identity and simplfy:

$\frac{1}{\cos x \sin x} - \frac{\cos^{2} x}{\cos x \sin x} = \frac{1 - \cos^{2} x}{\cos x \sin x} = \frac{\sin^{2} x}{\cos x \sin x} = \frac{\sin x}{\cos x} = \tan x = \tan x$"

My questions:

are associated with, the part of, the question in quotations. First question, is how does $\frac{\csc x \sin x}{\cos x \sin x} = \frac{1}{\cos x \sin x}$? I'm assuming everytime, you, have to simplfy and see the fraction, you get $\frac{1}{\cos x \sin x}$?

My other question, is how does $\frac{\sin x}{\cos x} = \tan x$? I'm assuming the fraction have always equaled $\tan x$? If someone could, explain my four questions it would be very helpful, for future problems.

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    $\begingroup$ How do you define $\tan$ and $\csc$? $\endgroup$ – Henrik Dec 20 '16 at 21:13
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Ah, you are trying to work trig identities before you are familiar with the definitions of the trig functions.

If you consider $\sin x$ and $\cos x$ as fundamental (although as you already know, $\sin^2x + \cos^2 x = 1$), then by definition

$$ \tan x = \frac{\sin x}{\cos x} \\ \cot x = \frac{\cos x}{\sin x} \\ \sec x = \frac1{\cos x}\\ \csc x = \frac1{\sin x} $$

So yes, you are allowed to freely use any of these four relationships (your problem used the first and the fourth).

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  • $\begingroup$ Thank you, I forgot about trig definitions, and, you showed me two new ones. $\endgroup$ – William Zlacki Dec 20 '16 at 21:16
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    $\begingroup$ That is not 1. And you did not use that as 1, you used it as $\frac1{\cos x \sin x}$ $\endgroup$ – Mark Fischler Dec 20 '16 at 21:30
  • $\begingroup$ Never mind, got it. $\endgroup$ – William Zlacki Dec 20 '16 at 21:43
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Since $LHS=RHS$. It's an identity...

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  • $\begingroup$ It would be even better typed it out in mathjax $\endgroup$ – suomynonA Feb 1 '17 at 4:32

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