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If I have $$ -\frac{\sin t}{\cos t} \div (1 - \sec^2t)$$ how do I simplify the divisor?

Let's say that I rewrite it as $$1 - \sec^2t = 1 - \frac{1}{\cos^2t}$$

Can someone detail the exact algebraic steps there? My basic question is:

Can I just say the following? $$ -\frac{\sin t}{\cos t} \div \left(1 - \frac{1}{\cos^2t}\right) = -\frac{\sin t}{\cos t} \times (1 - \cos^2t)$$

I suppose somebody didn't teach me fraction algebra like this in middle school, and now I'm struggling on petty things like this.

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    $\begingroup$ Can you write $1 \div (1 - \frac{1}{x})$ as $(1-x)$? Same thing. $\endgroup$ – Math Lover Oct 24 '20 at 10:49
  • $\begingroup$ @MathLover I'm not sure your comment is helpful, because clearly they are not confident with working with fractions in this way. They may legitimately not know the answer to your question, and so be uncertain as to whether you are saying "of course you can do this", or "of course you cannot do this". $\endgroup$ – Morgan Rodgers Oct 24 '20 at 10:54
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    $\begingroup$ @MorganRodgers I am not sure I agree with you. I translated the problem from trigonometry to regular fraction thinking it may help the OP to think in the right direction. Of course there is an assumption that the person understands fraction in regular context and is just confused in this case. $\endgroup$ – Math Lover Oct 24 '20 at 10:58
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No.

$\begin{align}-\dfrac{\sin t}{\cos t}\div (1-\sec^2t) &= -\dfrac{\sin t}{\cos t}\div \left(1-\dfrac{1}{\cos^2t}\right) = -\dfrac{\sin t}{\cos t}\div \left(\dfrac{\cos^2t-1}{\cos^2t}\right) \\&=-\dfrac{\sin t}{\cos t}\times \left(\dfrac{\cos^2t}{\cos^2t-1}\right)=\dfrac{\sin t \cos t}{1-\cos^2t} =\dfrac{\sin t \cos t}{\sin^2t} \\&=\dfrac{\cos t}{\sin t}\end{align}$

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$$ - \frac{ \sin{t} }{ \cos{t} } \div (1 - \frac{1}{\cos^2{t}}) = \\ - \frac{ \sin{t} }{ \cos{t} } \div (\frac{\cos^2{t} - 1}{\cos^2{t}}) = \\ - \frac{ \sin{t} }{ \cos{t} } \div (\frac{- \sin^2{t}}{\cos^2{t}}) = \\- \frac{ \sin{t} }{ \cos{t} } \cdot (\frac{- \cos^2{t}}{\sin^2{t}}) = \\ =\frac{\cos{t}}{\sin{t}}$$

Using the fact that: $$-\sin^2{t} = \cos^2{t} - 1$$

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  • $\begingroup$ This doesn't at all answer the question they are asking. $\endgroup$ – Morgan Rodgers Oct 24 '20 at 10:55
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    $\begingroup$ @MorganRodgers Well, he asked how does he simplify the divisor which is really the main idea in this answer - as an example. And if you are talking about the fact I didn't specify "yes" / "no" - then, sometimes math can explain better than words, just need to read between the lines. $\endgroup$ – WATER Oct 24 '20 at 10:56
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Let $$M=1-\frac{1}{\cos^22t}$$ and $$N=1-\cos^22t.$$ The question is asking whether $M$ and $N$ are multiplicative inverses, or in other words whether $MN=1$ in most interesting cases. Let's check. We find that $MN$ is equal to

$$\left(1-\frac{1}{\cos^22t}\right)(1-\cos^22t)=2-\left(\cos^22t+\frac{1}{\cos^22t}\right),$$ which is clearly not identically equal to $1,$ which puts the answer to your question in the negative.

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Note that

$$1-\frac1x = 1-x \iff \frac1x=x \iff x^2=1$$

that is by $x=\cos^2t$

$$\cos^4t=1 \iff t=k\pi$$

which is not true in general and therefore your step is not correct.

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No.

$\lvert \sec x\rvert \ge 1$, so $1-\sec^2 x \le 0$, so division by $1-\sec^2 x$ flips the sign (if the division is defined).

$0 \le \lvert \cos x\rvert \le 1$, so $0\le 1 - \cos ^2 x \le 1$, so multiplication by $1-\cos ^2 x$ keeps the sign unchanged.

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