## Identify useful patterns and use methods to solve in a few steps

We will take up in this session two selected NCERT Trigonometry problems which we will solve in a few steps. In the process, we will **highlight the problem solving approach**, in which **hidden useful patterns will be identified and quick simplification achieved using useful patterns hidden in the problem**.

We may classify these as **elegant within a minute problems**, where conventional deductions may take a backseat leaving the lead to pattern and method based high speed problem solving.

Generally these **problems are solved in mind** with minimum writing. In these solutions **number of steps are reduced significantly** so that even with writing, the approach saves valuable time.

We will provide explanations along with solutions.

Let us start solving now.

### Solutions to NCERT level Trigonometry problems—set 1

#### Problem 1.

Find the value of $(1+\tan \theta +\sec \theta)(1+\text{cot } \theta - \text{cosec } \theta)$ among the following choices. Justify your answer.

- $0$
- $1$
- $2$
- $-1$

#### Solution 1.

In these cases of *product of two trigonometric expressions*, the **time taking approach is to actually proceed with expanding the multiplication**. This would result in many terms. Hopefully we will then find cancellations and carry out simplifications to reach the simple solution.

Generally looking deeper,

We find possibilities that

make the two factors much easier to handle so that after multiplication of the two transformed expressions, number of terms are minimized.This is ourfirst objective.

In this case then, first question we ask is, **how can we make the terms of the two expressions more similar so that we can use the minus sign in the second expression and minimize the number of terms in the product?**

In two ways we can do it, either take $\tan \theta$ **factored out** of first expression, or take $\text{cot } \theta$ out of second expression. We will take the first path,

Given expression,

$E=(1+\tan \theta +\sec \theta)(1+\text{cot } \theta - \text{cosec } \theta)$

$=\tan \theta \left(\text{cot } \theta +1+\displaystyle\frac{\sec \theta}{\tan \theta}\right)(1+\text{cot } \theta-\text{cosec } \theta)$

$=\tan \theta(1+\text{cot } \theta + \text{cosec } \theta)(1+\text{cot } \theta - \text{cosec } \theta)$

$=\tan \theta \left[(1+\text{cot } \theta)^2 - \text{cosec}^2 \theta \right]$

$=\tan \theta \left[1+2\text{cot } \theta - (\text{cosec}^2 \theta - \text{cot} ^2 \theta)\right]$

$=\tan \theta.2\text{cot } \theta$, $(\text{cosec}^2 \theta - \text{cot} ^2 \theta)=1$ and it cancels out with the other 1,

$=2$.

**Answer:** Option c: 2.

The **pattern we detected** was,

The nearly similar terms in the two expressions and the opposite signs, opening up the possibility to use, $(a+b)(a-b)=a^2 - b^2$.

To make the pattern usable we factored $\tan \theta$ out of the first expression. This is the **method** we call **Term factoring out**. It is a general algebraic minor method that we use extensively in Algebra, Surds and Trigonometry. Here we can call it specifically **Trigonometric term factoring out**.

You may ask, how do I know that $\tan \theta$ factoring will make the two expressions similar and we will be able to use $(a+b)(a-b)=a^2 - b^2$?

#### How it was decided that $\tan \theta$ need to be factored out

Let us try to produce the action steps and reasoning,

- First, we analyzed the expressions, noticed two factors, decided from experience to try simplifying the factors first before multiplication so that
**number of terms are reduced as much as possible**. This is**formation of clear objective**.*While solving any such problem this objective will work.* - Second, on closer look detected the similarity of the terms of the two factors with a significant difference of minus sign in the second factor. With this fact in mind, next step was automatic.
- Third, formation of specific objective—how can we make the two expressions more similar and use the minus sign?
- Fourth, as a
**first trial****factored out $\tan \theta$ from****first two terms of the first expression**and found that the terms became exactly same as the first two terms of the second expression. Going ahead with the third term was natural.

Mark the phrase, **first trial**. Factoring out $\tan \theta$ has been a trial, but a trial based on experience and promise, and made after analyzing the problem. Trials we make often in problem solving, but none of the trials is random, and comes at a later stage of problem solving after analysis. In most cases, this type of trial we carry out with near certainty borne out of experience and deductive reasoning.

**A truth** which **aids good decision making** in math problem solving is,

The solving of the problem involves simplification from a complex two expression factors to a numeric result. So there must be patterns hidden in the expressions that if used will lead to simplification, not more complexity.

In any case, most such trials need a few seconds at most to carry out in mind, so we always explore possibilities like this.

The **second truth** is,

Unless you explore new possibilities, hidden useful patterns, how can you ever find them?

If there were no writing requirement, because of the **inherent pattern based simplicity of the solution steps**, the problem could have been solved wholly in mind without any writing, well within a minute.

#### Problem 2

Find the value of $(\sec A+\tan A)(1-\sin A)$ among the following choices. Justify your answer.

- $\sec A$
- $\sin A$
- $\text{cosec } A$
- $\cos A$

#### Solution 2

Here we will take a different approach. Observing $(1-\sin A)$ we explore the possibility of converting to $(1-\sin A)(1+\sin A)=1-\sin^2 A=\cos^2 A$, by multiplying and dividing by $(1+\sin A)$. We could see the solution immediately and follow the path without hesitation.

Let us explain,

$(\sec A+\tan A)(1-\sin A)\times{\displaystyle\frac{1+\sin A}{1+\sin A}}$

$=\displaystyle\frac{(\sec A+\tan A)\cos^2 A}{1+\sin A}$

$=\displaystyle\frac{\cos A +\sin A.\cos A}{1+\sin A}$

$=\cos A$.

**Answer.** Option d: $\cos A$.

#### Action steps and reasoning of problem solving

In this problem we have used a **different type of pattern and method.**

Let us first list out the pairs of trigonometric functions that we call, **friendly trigonometric functions pairs**. These are,

- $1 \pm \sin A$ or $1 \pm \cos A$. These use the most popular trigonometric identity, $\sin^2 A + \cos^2 A=1$.
- $\sec A \pm \tan A$. These use the highly effective group of identities of the form, $\sec^2 A - \tan^2 A=1$, Or, $\sec A - \tan A=\displaystyle\frac{1}{\sec A + \tan A}$, and
- $\text{cosec } A \pm \text{cot } A$. These use the group of highly effective identities, $\text{cosec}^2 A - \text{cot}^2 A=1$, or, $\text{cosec } A - \text{cot } A=\displaystyle\frac{1}{\text{cosec } A + \text{cot } A}$.

In **trigonometric expression simplifications** these standard and well known identities form our basic trigonometric problem solving resource just like arrows in the quiver of arrows of a hunter. The more such arrows you can form and put in your problem solving quiver of arrows, faster you can solve, and solve a wider range of problems.

Whenever we encounter any such expression, we make quick trials mentally, exploring simplifying possibilities. We knew that using the other factor $\sec A+\tan A$ also solution can be reached quickly, but we decided for $1-\sin A$, as $sin$ function is simpler than $sec$ or $tan$. $sin$ and $cos$ are the first level functions, the other two arise after specific operations involving first level functions. Thus if** we detect same promise in two expressions we choose the simpler one.** The rest was algebraic and straightforward.