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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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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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