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The questions asks for $$\frac{1}{\sin x\cos x} - \frac{1}{\tan x}$$ to be written in terms of sin and cosine

I tried it many different ways and thought the awswer would be $$\frac{1}{\sin x\cos x} - \frac{1}{\frac{\sin x}{\cos x}}$$

But this awnser doesn't seem the be the right one! If someone could explain how to write the equation correctly in terms of sin and cos it would be appreciated!

ps: The first question asked for $\cos x\tan x$ to be simplified and the right answer was $$\cos x\left(\frac{\sin x}{\cos x}\right)$$

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    $\begingroup$ To be fair, you have translated into sines and cosines only. It's just that you're expected to follow the "give a mouse a cookie" impulse: Well, nobody likes "nested" fractions, so let's clean that up. And now that I look at it, it sure would be nice to combine those two. And oh look! $1 - cos^2 x$ showed up, I know that guy... (but this story has a reasonable ending). $\endgroup$ – pjs36 Jun 12 '17 at 3:29
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Jun 12 '17 at 9:39
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One has $$\frac1{\sin x\cos x}-\frac1{\tan x} =\frac1{\sin x\cos x}-\frac{\cos x}{\sin x} =\frac{1-\cos^2x}{\sin x\cos x} =\frac{\sin^2x}{\sin x\cos x}=\frac{\sin x}{\cos x}=\tan x.$$

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  • $\begingroup$ Thanks so much for the work and the answer! Really helped me understand and tanx was the right solution! $\endgroup$ – Joseph Burrows Jun 12 '17 at 3:42
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$\displaystyle \frac{1}{\sin(x)\cos(x)}-\frac{1}{\tan(x)}=\frac{1}{\sin(x)\cos(x)}-\frac{\cos(x)}{\sin(x)}=\frac{1}{\sin(x)\cos(x)}-\frac{\cos^2x}{\sin(x)\cos(x)}=\frac{1-\cos^2(x)}{\sin(x)\cos(x)}=\frac{\sin^2(x)}{\sin(x)\cos(x)}=\boxed{\frac{\sin(x)}{\cos(x)}}$

P.S. For your first problem, $\displaystyle \cos(x)\tan(x)=\cos(x)\frac{\sin(x)}{\cos(x)}=\sin(x)$

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Or:

$$\frac{1}{\sin x\cos x} - \frac{1}{\tan x}=\frac{\sin^2x+\cos^2x}{\sin x\cos x} - \frac{1}{\tan x}=\frac{\sin x}{\cos x} + \require{cancel}\cancel{\frac{\cos x}{\sin x}}- \cancel{\frac{1}{\tan x}}=\tan x.$$

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