The equation will simplify to
\begin{align} & = \sqrt{\cos^2 100^\circ}\cdot \sec100^\circ \\[8pt] & = \cos100^\circ\cdot\sec100^\circ \\[8pt] & = 1 \end{align}
But the answer key says that the correct answer is $-1$?
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The answer key is correct; the cosine of $100^{\circ}$ is negative. – egreg Apr 22 '14 at 16:46
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@egreg but so is secaant of 100∘, so wouldn't they completely cancel out leaving the answer to be 1? – Niharika Apr 22 '14 at 16:50
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@egreg,I am not sure why that should matter.$\cos$ and $\sec$ both have the same sign. – rah4927 Apr 22 '14 at 16:50
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@rah4927 The fact is that $\cos100\ne\sqrt{1-\sin^2100}$. Just so simple. – egreg Apr 22 '14 at 16:53
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What is $\sqrt{(-1)^2}$? – evil999man Apr 22 '14 at 17:00
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@egreg,ah,so since $\cos100$ is negative,and the the square root gives us a non-negative value,we must have $-\cos100$ . – rah4927 Apr 22 '14 at 17:02
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@rah4927 yes except that you used the two times. – evil999man Apr 22 '14 at 18:25
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@Awesome,I don't get the second part of your comment. – rah4927 Apr 22 '14 at 18:32
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@rah4927 read your previous comment. – evil999man Apr 22 '14 at 18:33
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@Awesome,sorry I am still unable to see what I used two times. – rah4927 Apr 22 '14 at 18:37
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@rah4927 read after "negative, and" – evil999man Apr 23 '14 at 03:35
3 Answers
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We know that $\sqrt{x^2}=|x|$. So $\cos100$ will be negative because in the second quadrant. So $\sqrt{\cos^2100}=-\cos100$.
Satvik Mashkaria
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Root of $\cos(100)^2$ is always positive. So root of $\cos(100)^2$ is also positive. And the value of root of $1-\sin(100)^2$ will be $+\cos100$. Now $\sec100$ is negative, so the value is $1/-\cos100$. So $\cos100$ and $-\cos100$ get cancelled and we get -1 as the answer.
Glorfindel
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