Question

Make b the subject of formula: $ a=\dfrac{1+{{b}^{2}}}{1-{{b}^{2}}} $ .
A. $ b=\sqrt{\dfrac{a+1}{a-1}} $
B. $ b=\sqrt{\dfrac{a-1}{a+1}} $
C. $ b=2\sqrt{\dfrac{a-1}{a+1}} $
D. $ b=2\sqrt{\dfrac{a+1}{a-1}} $

Answer 1

Hint: We first explain the expression of the function. We convert the value of b with respect to a. The inverse function on being conjugated gives the value of $ b $ . At the end we interchange the terms to make it a general equation.

Complete step-by-step answer:
We need to find the inverse of the relation of $ a=\dfrac{1+{{b}^{2}}}{1-{{b}^{2}}} $ .
The given relation is of a expressed with respect to b.
If we take the inverse of the equation, we will get the value of b expressed with respect to a.
We first use componendo-dividendo formula for $ a=\dfrac{1+{{b}^{2}}}{1-{{b}^{2}}} $ to get
 $ \dfrac{a-1}{a+1}=\dfrac{1+{{b}^{2}}-1+{{b}^{2}}}{1+{{b}^{2}}+1-{{b}^{2}}}={{b}^{2}} $ .
Now we take the square root of the equation $ {{b}^{2}}=\dfrac{a-1}{a+1} $ .
We get $ b=\pm \sqrt{\dfrac{a-1}{a+1}} $ .
We expressed the relation of b expressed with respect to a.
From the given options we can see that the equation $ b=\sqrt{\dfrac{a-1}{a+1}} $ satisfies.
Therefore, the correct option is B.
So, the correct answer is “Option B”.

Note: We can verify the result by taking the composite function.
Putting the value of $ b=\sqrt{\dfrac{a-1}{a+1}} $ , in the equation of $ a=\dfrac{1+{{b}^{2}}}{1-{{b}^{2}}} $ we get
\[a=\dfrac{1+{{\left( \sqrt{\dfrac{a-1}{a+1}} \right)}^{2}}}{1-{{\left( \sqrt{\dfrac{a-1}{a+1}} \right)}^{2}}}=\dfrac{1+\dfrac{a-1}{a+1}}{1-\dfrac{a-1}{a+1}}=\dfrac{a+1+a-1}{a+1-a+1}=\dfrac{2a}{2}\].
From the above relation we get back a. this equation satisfies each other.