Componendo dividendo explained

Amazingly powerful Componendo dividendo

componendo dividendo

We have seen Componendo dividendo to simplify competitive math problems in topics as diverse as—number system, ratio and proportion, surds, trigonometry or algebra. Often it plays a key role in solving the problem quickly.

Componendo dividendo is powerful because,

  1. It simplifies an apparently complex expression greatly.
  2. Because of the special form of expression on which Componendo dividendo can be applied, it is easy to identify where it can be applied,
  3. It is easy to understand and use in application. It is just a three-step simple method,
  4. Due to its simplicity, the three step process can be done in mind and so we can consider the ability to apply Componendo dividendo under mental math skills. This is especially effective for answering competitive math questions where you need to select one of multiple choice answers quickly and don’t need to show the deductions.

At the basic level, Componendo dividendo is applied in two ways. We will explain with examples.

Componendo dividendo on $\displaystyle\frac{x+1}{x-1}$

Problem example 1.

What is the value of $x$ when $\displaystyle\frac{x+1}{x-1}=5$?

This is a standard form of expression where immediate application of componendo dividendo will give you the answer in a few seconds without need to write any calculation steps on paper.

The three steps in componendo dividendo to solve the problem are:

Step 1: Add 1 to both sides of the equation to reduce the numerator of the LHS to a single term $2x$ without changing the denominator $x-1$,

$\displaystyle\frac{x+1}{x-1}+1=5+1$,

Or, $\displaystyle\frac{2x}{x-1}=6$.

Step 2: Subtract 1 from both sides of the equation to reduce the numerator of LHS to 2 (x cancels out) without changing the denominator $x-1$,

$\displaystyle\frac{x+1}{x-1}-1=5-1$,

Or, $\displaystyle\frac{2}{x-1}=4$.

Step 3: Now divide result from step 1 by result from step 2 to eliminate the common denominator, $x-1$,

$x=\displaystyle\frac{3}{2}$ .

We can arrive at the same result and verify whether our result is correct by conventional method of cross-multiplication.

Conventional method

Problem: What is the value of $x$ when $\displaystyle\frac{x+1}{x-1}=5$?

Cross-multiplying,

$x+1=5x-5$,

Or, $4x=6$,

Or, $x=\displaystyle\frac{6}{4}=\frac{2}{3}$.

In the second basic form, the subtractive expression $x-1$ appears in the numerator instead of the denominator. In this form, at the second step you need to subtract the two sides of the equation from 1, instead of subtracting 1 from two sides of the equation.

Componendo dividendo on $\displaystyle\frac{x-1}{x+1}$

Problem example 2.

What is the value of $x$ when $\displaystyle\frac{x-1}{x+1}=\displaystyle\frac{2}{7}$?

The main difference with the first form is, the subtractive $x-1$ is in the numerator instead of the denominator in the first case. Due to this difference there will be a basic difference in the second of the three steps,

Step 1: Add 1 to both sides of the equation to reduce the numerator of the LHS to a single term $2x$ without changing the denominator $x+1$.

$1+\displaystyle\frac{x-1}{x+1}=1+\displaystyle\frac{2}{7}$,

Or, $\displaystyle\frac{2x}{x+1}= \displaystyle\frac{9}{7}$.

Step 2: Subtract both sides of the equation from 1 to reduce the numerator of the LHS to 2 (x cancels out) without changing the denominator $x+1$.

$1-\displaystyle\frac{x-1}{x+1}= 1- \displaystyle\frac{2}{7}$,

Or, $\displaystyle\frac{2}{x+1}= \displaystyle\frac{5}{7}$.

Step 3: Now divide result from step 1 by result from step 2 to eliminate the common denominator, $x+1$.

$x=\displaystyle\frac{9}{5}$ .

You should be careful in identifying in what form the expression appears, whether the subtractive $x-1$ is in the numerator or in the denominator. If it is in the numerator, you need to subtract the two sides of the equation from 1 at the second step, so that the resultant numerator stays positive.

Instead of a single variable, the expression to be simplified can appear in two variables. Let us consider the third general form.

Componendo dividendo on $\displaystyle\frac{x+y}{x-y}$

Problem example 3.

What is the value of $x:y$ when $\displaystyle\frac{x+y}{x-y}=9$?

In this case we have one equation and two variables. So we won’t get actual values of two variables. At most we may get the ratio. That is the simplest result we usually need under the circumstances.

Step 1. Adding 1 to both sides of the equation,

$\displaystyle\frac{x+y}{x-y}+1=9+1$,

Or, $\displaystyle\frac{2x}{x-y}=10$

Step 2. Subtracting 1 from two sides of the equation,

$\displaystyle\frac{x+y}{x-y}-1=9-1$,

Or, $\displaystyle\frac{2y}{x-y}=8$.

Step 3. Dividing the result of Step 1 by result of Step 2 to eliminate common denominator $x-y$,

Taking ratio, $\displaystyle\frac{x }{ y}=\frac{5}{4}$,

Or, $x:y=5:4$.

Even in the simple form, componendo dividendo is the preferred and quicker way to reach the result for us.

To show how this highly popular three-step simple method can help to solve a relatively more complex problem quickly, we will use a fourth problem example from Trigonometry.

Problem example 4. Componendo dividendo in Trigonometry

If $a=\sec \theta+\tan \theta$, then $\displaystyle\frac{a^2-1}{a^2+1}$ is,

  1. $\sec \theta$
  2. $\cos \theta$
  3. $\tan \theta$
  4. $\sin \theta$

Solution to Problem example 4. Problem analysis

The target expression is in perfect form for applying Componendo and Dividendo only if instead of $a$, we were given the value of $a^2$.

So instead of thinking of applying Componendo Dividendo, now we look for ways to transform the input value of $a$ to $a^2$. This is what we call Analytical approach example in which to get the given target we determine how we need to transform given input. Our immediate action changes accordingly.

Solution to Problem example 4. Problem solving execution

We achieve this immediate goal by using a special property of what we call one of the golden trigonometric function pairs, that is, $sec \theta + tan \theta$. The special property of this friendly trigonometric function pair is,

$sec \theta + tan \theta = \displaystyle\frac{1}{sec \theta - tan \theta}=a$, in this case.

The property is true because, $sec^2 \theta - tan^2 \theta=1$.

Thus we get the value of $a^2$,

$a^2=\displaystyle\frac{sec \theta + tan \theta}{sec \theta - tan \theta}$.

And finally, applying the componendo dividendo on the equation (subtracting 1 from both sides, again adding 1 to both sides of the original equation and taking the ratio of the two),

$\displaystyle\frac{a^2-1}{a^2+1}=\frac{tan\theta}{sec\theta}=sin\theta$.

Answer: Option d: $sin \theta$.

Just remember, in solving tough problems, you need to think analytically more rather than just apply a method mechanically.

We have applied Componendo Dividendo at the end, and that end we had visualized at the beginning by our initial problem analysis and change of immediate goal to getting $a^2$ from given value of $a$.

As soon as this changed goal we had achieved by applying an especially powerful property of the golden function pair $sec \theta+tan \theta$, we practically had our solution in the pocket.

Though it was difficult to imagine at the beginning, the problem could be solved wholly in mind without writing any bit on paper, which is the ideal situation for answering competitive test questions. This falls under the skill-set of mental maths, that is, how much you can analyze and process in mind easily and quickly.


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