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If someone could help with either of these problems that would be awesome!

$(\tan x)^2 \leq |1 - 2(\cot x)^2|$

$x^{\sin(x-a)}>1$ where $0< x < \frac{\pi}{2}$ and $a>0$

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Please tag as (homework). – Did Mar 21 '11 at 8:53
@Didier: How do you know this is homework? The (homework) tag is a voluntary statement by the original poster that a question is homework. – Isaac Mar 21 '11 at 20:12
It is homework, but I'm more interested in understanding how to do it than just having the answers... – Shawn Mar 21 '11 at 20:18
@Isaac Some of us know it when they see it. – Did Mar 23 '11 at 7:15
The (homework) tag is not related to the OP being more or less interested in understanding a method than in getting a flat answer (eventhough my unmitigated preference goes to the former rather than the latter). This tag indicates, well... homework. – Did Mar 23 '11 at 7:32

2 Answers 2

Hint: For the second, if $x \gt 1$ and it is raised to any positive power...

Then there are three more versions of the above.

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i'm not sure i follow. if x>1 and it is raised to any positive than the result must be >1, what conditions can you state that ensure that the power sin(x-a) is positive given that a is any positive.... doesn't that imply that that power can be negative no matter what x value >1 you choose? – Shawn Mar 21 '11 at 20:02
Over what range of $\theta$ is $\sin \theta \gt 0$? Given $a$, can you express the restrictions on $x$ to satisfy this? You might start with the specific cases $a=0, 1/4$ and graph $\sin (x-a)$ as a function of $x$ to see what happens. – Ross Millikan Mar 21 '11 at 20:27
well I can see that you can describe x in terms of a for x>1 with a<x<a+pi, but I was under the impression that to solve the inequality you must have some x that works for all values of a, not to give an answer in terms of a. maybe i am misinterpreting? – Shawn Mar 21 '11 at 20:35
No, the allowable $x$ will depend on $a$. You are given that $x\lt \pi/2$, so one part of the solution is $1 \lt x \lt \pi/2, 0\lt a \lt x$. You are basically mapping out the regions in the $(x,a)$ plane where the inequality is satisfied. There is also a part with $x \lt 1$ and $a$ can be larger than $x$ even with $x \gt 1$ How is that? – Ross Millikan Mar 21 '11 at 22:25

The first one, rewrite cotangent in terms of tangent and apply the ideas suggested in the answers to your previous question. The second one, I'm not entirely sure what you intended to write.

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