#5**+5 **

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

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.

Bertie May 6, 2014

#1**0 **

atan(1/sqrt(x))-atan(sqrt(x)) = 105

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\sqrt{{\mathtt{x}}}}}}\right)}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\sqrt{{\mathtt{x}}}}\right)} = {\mathtt{105}} \Rightarrow \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\sqrt{{\mathtt{x}}}}\right)} = {\frac{\left({\mathtt{12}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{2\pi}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\sqrt{{\mathtt{x}}}}}}\right)}{\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{{\mathtt{12}}}}$$

Umm, that doesn't look very helpful does it!

Melody May 6, 2014

#2**0 **

arctan(1/√x) and arctan(√x) are angles.

If you take the tangent of the equation and use the usual identity

tan(A - B) = (tan A - tan B)/(1 + tan A.tan B)

you finish up with a quadratic in √x.

I don't have time to type it in at the moment, I'll do so later if it hasn't been done in the meantime.

Also, is that 105 degrees or radians ? My guess is degrees.

Bertie May 6, 2014

#4**0 **

Hallo

$$\alpha =atan(\frac{1}{\sqrt{x}})=atan(\frac{1}{u})$$

$$\beta =atan(\sqrt{x})==atan(u)$$

$$\alpha + \beta \quad must \quad be \quad 90\;degrees!$$

In a right-angled triangle with "u" one Side and "1" another Side

$$\tan(\alpha)=\frac{1}{u}$$ and $$\tan(\beta)=\frac{u}{1}$$

$$atan(\tan(\alpha))=atan(\frac{1}{u})=\alpha$$

$$atan(\tan(\beta))=atan(\frac{u}{1})=\beta$$

In a right-angled triangle $$\alpha + \beta = 90\;degrees$$

$$\alpha - \beta \quad cannot \quad be \quad greater \quad than \quad 90\; degrees.$$

.Guest May 6, 2014

#5**+5 **

Best Answer

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

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.

Bertie May 6, 2014