How to solve $\arctan{\sqrt{\frac{10-x}{x}} + \arccos\sqrt{\frac{10-x}{10}}}= \frac\pi{2}$ Can you tell me if it is possible to solve $$\arctan{\sqrt{\frac{10-x}{x}} + \arccos\sqrt{\frac{10-x}{10}}}= \frac\pi{2}\ ?$$
wolfram doesn't say it is not computable, but says there are no real solutions, whereas we know the solution is $x=1$
Any hints on how to proceed?
Thanks
 A: Actually, $1$ is not the only solution. Every $x$ such that $0< x\leq10$ is a solution. To see this, observe that the equation models a right angle triangle with legs $\sqrt x$ and $\sqrt{10-x}$.
A: Let $y=\sqrt{\frac{10-x}{x}}$, let $z=\sqrt{\frac{10-x}{10}}$. Take the cosine of both sides:
$$\cos\left(\arctan y+\arccos z\right)=0.$$
Apply the angle addition formula.
$$\cos(\arctan y)z-\sin(\arctan y)\sin(\arccos z)=0.$$
From the definition of trigonometric ratios, we know that $\cos(\arctan y)=\frac{1}{\sqrt{y^2+1}}$, $\sin(\arctan y)=\frac{y}{\sqrt{y^2+1}}$, and $\sin(\arccos z)=\sqrt{1-z^2}$. This leaves us with
$$\frac{z}{\sqrt{y^2+1}}-\frac{y}{\sqrt{y^2+1}}\sqrt{1-z^2}=0$$
$$z-y\sqrt{1-z^2}=0$$
$$\sqrt{\frac{10-x}{10}}-\sqrt{\frac{10-x}{x}}\sqrt{1-\frac{10-x}{10}}=0$$
$$\sqrt{\frac{10-x}{10}}-\sqrt{\frac{10-x}{x}}\sqrt{\frac{x}{10}}=0$$
$$0=0.$$
This equation boils down to a tautology, so every value of $x$ is a solution.
A: Let $\arctan\sqrt{\dfrac{10-x}x}=y\  \ \ \ (1)$
$\implies 0\le y\le\dfrac\pi2$ as $\sqrt{\dfrac{10-x}x}\ge0$
$\dfrac{10-x}x=\tan^2y\iff x=10\cos^2y$
$\implies\sqrt{\dfrac{10-x}{10}}=+\sin y$ as $0\le y\le\dfrac\pi2$
$\implies\arccos\sqrt{\dfrac{10-x}{10}}=\arccos(\sin y)=\dfrac\pi2-\arcsin(\sin y)$
$\implies\arccos\sqrt{\dfrac{10-x}{10}}=\dfrac\pi2-y\  \ \ \ (2)$ as $0\le y\le\dfrac\pi2$
Use $(1),(2)$
