$$\tan^{-1}(\frac{1}{\sqrt{x}})-\tan^{-1}(\sqrt{x})=105,$$

so, taking the tangent of both sides, and, using the identity

$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B},$$

together with

$$\tan(\tan^{-1}(\frac{1}{\sqrt{x}}))=\frac{1}{\sqrt{x}}\quad \text{ and }\quad \tan(\tan^{-1}(\sqrt{x}}))=\sqrt{x},$$

we have

$$\frac{1}{\sqrt{x}}-\sqrt{x}=2\tan 105.$$

Multiplying throughout by √x and rearranging,

$$x+2\sqrt{x}\tan 105-1=0,$$

which is a quadratic in √x.

Solving that, and taking the positive root gets √x≈7.595754.

To show that this satifies the original equation, remember that arctan is multivalued.

$$\tan^{-1}(\frac{1}{\sqrt{x}})-\tan^{-1}(\sqrt{x})=(7.5+k_{1}180)-(82.5+k_{2}180),$$

where k_{1} and k_{2} are integers.

k_{1} = 1 and k_{2 }= 0 produces the result 105, but there are an infinite number of other possibles.