B Given an equation, find the value of this expression

[littlemathquark]

TL;DR Summary

İf $-1+\dfrac{2}{1-x}=\sqrt 3$ then find $\dfrac{2x}{1-x^2}\cdot \sqrt 3$

İf $-1+\dfrac{2}{1-x}=\sqrt 3$ then find $\dfrac{2x}{1-x^2}\cdot \sqrt 3$

I can solve it with two ways:
1) directly by finding $x$
2) Let $x=\tan\alpha$ and go on and answer is $1.$

But I need another way. Are there an another way?

 

[Mark44]

I can solve it with two ways:
1) directly by finding $x$
2) Let $x=\tan\alpha$ and go on and answer is $1.$

But I need another way. Are there an another way?

How are you getting an "answer" of 1? This can't be the value of x as it would make the equation undefined. The value I get for x doesn't give a value of 1 for the expression $\frac{2x}{1 - x^2}\sqrt 3$.

 

[littlemathquark]

How are you getting an "answer" of 1? This can't be the value of x as it would make the equation undefined. The value I get for x doesn't give a value of 1 for the expression $\frac{2x}{1 - x^2}\sqrt 3$.

I mean $\frac{2x}{1 - x^2}\sqrt 3=1$

 

[fresh_42]

TL;DR Summary: İf $-1+\dfrac{2}{1-x}=\sqrt 3$ then find $\dfrac{2x}{1-x^2}\cdot \sqrt 3$

But I need another way.

a) Why?
b) Explain the techniques that qualify for "another".

 

[littlemathquark]

a) Why?
b) Explain the techniques that qualify for "another".

a) Just curious
b) I don't know the techniques that qualify as "another"

 

[fresh_42]

a) Just curious
b) I don't know the techniques that qualify as "another"

So let me make this clear: you have a precise definition of the value $x$ and want to find a value $g(x)$ without using $x$ and no idea what is allowed to be used instead?

 

[littlemathquark]

So let me make this clear: you have a precise definition of the value $x$ and want to find a value $g(x)$ without using $x$ and no idea what is allowed to be used instead?

I want to find value of g(x) using/manupulating given equation( as I said I found it using trigonometry but I wonder other ways) . I assume that not known precise value of x.

 

[Mark44]

I can solve it with two ways:
1) directly by finding $x$
2) Let $x=\tan\alpha$ and go on and answer is $1.$

I get a solution for x that checks out. If I substitute this value into $\frac{2x}{1 - x^2}\sqrt 3$, I don't get a result of 1 for that expression.

What did you get for x?
If you replace x in the expression above by $\tan(\alpha)$, the expression becomes $\frac{2\tan(\alpha)}{1 - \tan^2(\alpha)}\sqrt 3 = \sqrt 3 \tan(2\alpha)$. How does that simplify to 1?

 

[littlemathquark]

I get a solution for x that checks out. If I substitute this value into $\frac{2x}{1 - x^2}\sqrt 3$, I don't get a result of 1 for that expression.

What did you get for x?
If you replace x in the expression above by $\tan(\alpha)$, the expression becomes $\frac{2\tan(\alpha)}{1 - \tan^2(\alpha)}\sqrt 3 = \sqrt 3 \tan(2\alpha)$. How does that simplify to 1?

I found $x=2-\sqrt 3$ and evaluated it in $(2x/(1-x^2)).\sqrt 3$

 

[littlemathquark]

I found $x=2-\sqrt 3$ and evaluated it $(2x/(1-x^2)).\sqrt 3$

$(x+1)/(1-x)=\sqrt 3$ and Let $x=\tan\alpha$ so $(1+\tan\alpha)/(1-\tan\alpha)=\sqrt 3$ and $1=\tan 45$
$\tan(45+\alpha)=\sqrt 3=\tan 60$
$\alpha=15^{\circ}$

 

[Mark44]

I found $x=2-\sqrt 3$ and evaluated it in $(2x/(1-x^2)).\sqrt 3$

Rechecking my work, I get the same value for x as you did.

 

[anuttarasammyak]

\frac{2x}{1-x^2}=\frac{1}{1- x}-\frac{1}{1+x}with\frac{1+x}{1-x}=\sqrt{3}seems useful.

 

[littlemathquark]

\frac{2x}{1-x^2}=\frac{1}{1- x}-\frac{1}{1+x}with\frac{1+x}{1-x}=\sqrt{3}seems useful.

I can't use your hint. Can you give me more detail?

 

[anuttarasammyak]

\frac{2x}{1-x^2}=\frac{1}{1- x}-\frac{1}{1+x}=(1-\frac{1}{\sqrt{3}})\frac{1}{1-x}=(1-\frac{1}{\sqrt{3}})\frac{\sqrt{3}+1}{2}=\frac{1}{\sqrt{3}}

 

[littlemathquark]

Thank you. Now, I have three ways for solution the problem.

 

[anuttarasammyak]

Geometry of
2-\sqrt{3}=tan \frac{\pi}{12}

 

[littlemathquark]

Geometry of
2-\sqrt{3}=tan \frac{\pi}{12}
View attachment 358523

Thank you. Now, I have three ways for solution the problem.

Fourth solution:
$\dfrac{1+x}{1-x}=\sqrt3$ so $(1+x,1-x)=(\sqrt3k,k)$ and $k=\sqrt3-1$

$2x=(1+x)-(1-x)$ and $1-x^2=(1+x)(1-x)$ so $\dfrac{(\sqrt3-1)k}{\sqrt3k\cdot k}\cdot \sqrt 3=1$

 

[anuttarasammyak]

\frac{2\sqrt{3}x}{1-x^2}:=A
Ax^2+2\sqrt{3}x-A=0
x=\frac{-\sqrt{3}\pm\sqrt{A^2+3}}{A}=2-\sqrt{3}
We observe
A=1 with + sign satisfies the relation.