What is the Positive Integral solution of this expression?

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and how to do it?


closed as off-topic by kingW3, user223391, Daniel W. Farlow, Vladhagen, Shailesh Jan 22 '17 at 0:24

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  • $\begingroup$ You have two variables and one equation... $\endgroup$ – pie314271 Jan 21 '17 at 17:43
  • $\begingroup$ Is it an incomplete question? cuz I don't know, it is given in my book with no solution. only answer. and I can't handle it. $\endgroup$ – Pushkar Soni Jan 21 '17 at 17:47
  • $\begingroup$ That's odd.. it seems like an incomplete question unless you want one variable in terms of another. Or if there's another equation in x and y. $\endgroup$ – pie314271 Jan 21 '17 at 18:31
  • $\begingroup$ the answers in my book is :: for x=1, y=2 and for x=2, y=7. $\endgroup$ – Pushkar Soni Jan 21 '17 at 18:35

The way I've found to solve this equation is disentangling the expression by using some trigonometric identities.

First we rewrite the equation: $$x=tan\left(sin^{-1}\frac{3}{\sqrt{10}}-cos^{-1}\frac{y}{1+y^2}\right)$$

Now, we have in general $sin^{-1}(a)=tan^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)$ that leads to a more convenient way to write $sin^{-1}\left(\frac{3}{\sqrt{10}}\right)$. Being $a=\frac{3}{\sqrt{10}}$, you can operate to get $sin^{-1}\left(\frac{3}{\sqrt{10}}\right)=tan^{-1}(3)$

In a similar way we can get another tangent from the arccosine using the identity:


with $b=\left(\frac{y}{\sqrt{1+y^2}}\right)$ and operating we get $cos^{-1}\left(\frac{y}{\sqrt{1+y^2}}\right)=tan^{-1}\left(\frac{1}{y}\right)$. Then:


Now, using the identity for the tangent of the diffence: $$tan(c-d)=\frac{tan(c)-tan(d)}{1+tan(c)tan(d)}$$

We get:


It's easy to see that the pairs (1,2) and (2,7) are solutions. No more positive integer solutions are possible considering the meaning of our original equation: We calculate the tangent of an angle as the difference of approx. 71.57 degrees and some other angle always less than 90 degrees. There are only two angles with integer tangent between 0 and 71,57 degress.


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