$\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $

Asked May 19 '15 at 16:28

Active May 19 '15 at 16:57

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Express $$\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $$ as a rational integral equation between x and y.

This is what I've done:
Let $$t = \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}}$$
$$2\tan^{-1}(t/2) - \tan^{-1}t = \tan^{-1}\frac{t^3}{4+3t^2}$$
$$\implies \frac{\sqrt{1-x^2-y^2}}{y} = \frac{t^3}{4+3t^2}$$

Substituting the value of t and squaring both sides leads to very long calculation. Is there any other way to solve this problem?

edited May 19 '15 at 16:57

egreg

238,574

asked May 19 '15 at 16:28

Shaurya Gupta

4,339

This may help $$\cos^{-1} \frac{y}{\sqrt{1-x^2}}=2\cos ^{-1} \frac{\sqrt {3} }{2x}-\cos^{-1} \frac {x}{\sqrt{3}\times \sqrt{1-x^2}}$$ Hoping no careless mistake. – Someone May 19 '15 at 16:45
It is $$\color{red}{\cos ^{-1} \frac{2x}{\sqrt{3}}}$$ – Someone May 19 '15 at 16:53
@Mann Could you please describe in more detail how the above is useful? – Shaurya Gupta May 19 '15 at 16:59
@Mann Converting to $\cos^{-1}$ along with a bit of manipulation does reduce calculation but it's still quite a lot. – Shaurya Gupta May 19 '15 at 17:20
I believe the most compact relation between $x$ and $y$ is $y^2=3\left(x-\frac89x^3\right)^2$. – David H Jun 26 '15 at 14:46
@DavidH Yup, that's correct. – Shaurya Gupta Jun 27 '15 at 17:56

$\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $

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