Question

Define componendo and dividendo along with their applications in mathematics.

Answer 1

Hint: We first explain the componendo-dividendo formula. We convert the value of one variable with respect to the other. We express the concept with the use of examples.

Complete step-by-step answer:
The componendo-dividendo formula used for the equality of the fractions.
If we take $\dfrac{x}{y}=\dfrac{m}{n}$, then we can write it as $\dfrac{x+y}{x-y}=\dfrac{m+n}{m-n}$. The inverse of the relation gives $\dfrac{x-y}{x+y}=\dfrac{m-n}{m+n}$.
We try to understand the concept with the use of an example.
We need to find the inverse of the relation of $a=\dfrac{1+{{b}^{2}}}{1-{{b}^{2}}}$.
The given relation is of an 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.

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.