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Can anybody explain what $$\tan(\sin^{-1}(\frac x y))$$ equals?

I have to determine whether $$y'' \left(\tan \left(\sin^{-1}\left(\frac x y \right)\right) - \frac{x}{\sqrt{y^2-x^2}} \right)=0$$ is a linear differential equation. I thought that the left term might be equal to the right term, but I guess not.

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If you draw a right triangle with one angle $\theta$, the opposite side $x$, and the hypotenuse $y$ (which you know because of the $\arcsin$) you can find $\tan \theta=\frac x{\sqrt {y^2-x^2}}$

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  • $\begingroup$ so the whole term becomes 0? $\endgroup$ – Applied mathematician Nov 18 '12 at 17:05
  • $\begingroup$ Yes, that appears to be the case. $\endgroup$ – preferred_anon Nov 18 '12 at 17:34
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You can use trig identities. Let $y=\tan(\sin^{-1}(x))$, so $y^{2}+1=\sec^{2}(\sin^{-1}(x))$.
Further, since $\sec(x)=1/\cos(x)$, $\frac{1}{1+y^{2}}=\cos^{2}(\sin^{-1}(x))=1-\sin^{2}(\sin^{-1}(x))=1-x^{2}$, so $1+y^{2}=\frac{1}{1-x^{2}}$, and $y^{2}=\frac{x^{2}}{1-x^{2}}$ i.e.
$$\tan^{2}\left(\frac{x}{y}\right)=\frac{(x/y)^{2}}{1-(x/y)^{2}}=\frac{x^{2}}{y^{2}-x^{2}} \implies \tan\left(\frac{x}{y}\right)=\pm \frac{x}{\sqrt{y^{2}-x^{2}}}$$
I prefer the idea of thinking with triangles, but there's no harm in using identities!

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