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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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    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?2011-03-21
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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.