+0

# Solve for x

0
291
5

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

Guest May 6, 2014

#5
+889
+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 k1 and k2 are integers.

k1 = 1 and k2 = 0 produces the result 105, but there are an infinite number of other possibles.

Bertie  May 6, 2014
Sort:

#1
+91469
0

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

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

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

Melody  May 6, 2014
#2
+889
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
#3
+26406
0

Alan  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
+889
+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 k1 and k2 are integers.

k1 = 1 and k2 = 0 produces the result 105, but there are an infinite number of other possibles.

Bertie  May 6, 2014

### 12 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details