<s>[INST] 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
 [/INST] 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.
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