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

In this session, we will take up 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 **useful patterns and corresponding methods have been identified and applied for quick solution of the problems in a few steps**.

We may classify these as **elegant within a minute problems**, where instead of time-consuming conventional method, we adopt **pattern and method based high speed problem solving.**

Generally, these **problems are solved in mind** with minimum writing. In any case, 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 2

#### Problem 1.

Prove the identity,

$(\text{cosec } \theta - \text{cot } \theta)^2=\displaystyle\frac{1-\cos \theta}{1+\cos \theta}$

#### Solution 1. Identifying the key pattern and facing the first challenge

The **inviolable rule that we follow in proving identities** is to,

Apply operations on the LHS only to transform it to the RHS, that is, we move in a single direction, $\text{LHS} \Rightarrow \text{RHS}$.

We know it is always possible to expand the LHS and go through time-consuming deductive steps to arrive at the RHS. Instead of this longer path, we try to solve such problems in a few steps mentally. In the process, we look for and **find useful patterns and apply simplifying methods** to reach the solution quickly. This is what we call, **Problem solving in a few steps using patterns and methods.**

Generally the **useful pattern and effective method are lightly hidden in the problem**. Based on the nature of the problem, if you look for the **pattern-method pair**, you will certainly find it.

In this problem, the useful pattern that is directly visible is the **friendly trigonometric function pair**, $(\text{cosec } \theta - \text{cot } \theta)$. Generally we find this pattern to be used more effectively in its identity relation,

$\text{cosec } \theta - \text{cot } \theta=\displaystyle\frac{1}{\text{cosec } \theta + \text{cot } \theta}$.

This inverse relation between the additive-subtractive complementary function expressions, $(\text{cosec } \theta + \text{cot } \theta)$ and $(\text{cosec } \theta - \text{cot } \theta)$ is the outcome of the more well-known identity,

$(\text{cosec}^2 \theta - \text{cot}^2 \theta)=(\text{cosec } \theta + \text{cot } \theta)(\text{cosec } \theta - \text{cot } \theta)=1$.

The pattern is not hidden here, the suitable method to use the pattern is hidden. The useful pattern is of no use till you find the proper method to use it for quick solution of the problem.

In this case then, first question we ask is, **how can we use the friendly trigonometric function pair in the LHS expression effectively?**

To know the answer we **analyze our primary objective**. To solve the problem quickly, the easiest path must be, **making the LHS and RHS more similar in form**.

The **driving force here, is the objective**,

How can we use the pattern in making the LHS

more similarto the RHS?

Presently, the LHS and RHS are quite dissimilar in form.

Thus, **how to use the effective pattern** to make the **LHS more similar in form to the RHS** is the **main challenge** now.

#### Solution 1. Overcoming the first challenge by visualizing the square in the LHS as a product of two expressions

Can we look at the square expression in the LHS as a product of two same expressions? And then replace one of the factors by its complementary additive inverse of $(\text{cosec } \theta + \text{cot } \theta)$,

LHS,

$(\text{cosec } \theta - \text{cot } \theta)^2$

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

$=\displaystyle\frac{\text{cosec } \theta - \text{cot } \theta}{\text{cosec } \theta + \text{cot } \theta}$

The LHS and RHS are now very similar in form. The first challenge is overcome.

#### Solution 1. Final transformation

From this form of LHS, reaching the RHS is straightforward—just substitute,

$\text{cosec } \theta=\displaystyle\frac{1}{\sin \theta}$, and

$\text{cot } \theta=\displaystyle\frac{\cos \theta}{\sin \theta}$.

The common factor of $\sin \theta$ in the denominators of both terms of the numerator and denominator will cancel out leaving the RHS,

$\displaystyle\frac{\text{cosec } \theta - \text{cot } \theta}{\text{cosec } \theta + \text{cot } \theta}$

$=\displaystyle\frac{1-\cos \theta}{1+\cos \theta}$.

Provlem solved.

Though we have explained in great details, the operations are very easy once you see the possibility.

The key method has been to **visualize** the square expression, $(\text{cosec } \theta - \text{cot } \theta)^2$ as a product of two numbers of $(\text{cosec } \theta - \text{cot } \theta)$, and replace one of the factors with its equivalent inverse of the complementary expression, $\text{cosec } \theta - \text{cot } \theta=\displaystyle\frac{1}{\text{cosec } \theta + \text{cot } \theta}$.

In the **Pattern identification technique** here, we have identified the useful pattern and the suitable method by forming the **main objective of making the LHS more similar in form to the RHS**. This was **the driving force**.

By pattern identification technique we **identify the useful pattern as well as the suitable method** for achieving quick solution. The **pattern-method pair is important and useful**, not the pattern only.

#### Problem 2

Prove the identity,

$\displaystyle\frac{\cos \text{A}}{1+\sin \text{A}}+\displaystyle\frac{1 + \sin \text{A}}{\cos \text{A}}=2\sec \text{A}$

#### Solution 2: Deciding on the Primary and Secondary objectives

Here the form of the problem is different, we must combine the two LHS terms to reach the simplified form of RHS. This is the primary objective in this problem,

Combine the two LHS terms as easily as possible.

The main obstacle standing in the path of meeting the objective is, "dissimilar denominators" of the two LHS terms.

Immediately the question we ask,

How to make the two (or three for more fraction terms) denominators same?

This is the **secondary objective** in this case. If we can find a way to easily make the two denominators equal, it is more or less a certainty that the problem will be solved quickly.

#### Solution 2: Finding the pattern and suitable method to make the two denominators equal

Now we notice the presence of $(1+\sin \text{A})$ in the first term denominator, as well as the $\cos \text{A}$ in the numerator. This belongs to the class of friendly trigonometric function pairs, as we can just multiply the denominator and numerator by $(1-\sin \text{A})$, and transform the denominator to $\cos^2 \text{A}$.

Let us show the steps,

LHS,

$\displaystyle\frac{\cos \text{A}}{1+\sin \text{A}}+\displaystyle\frac{1 + \sin \text{A}}{\cos \text{A}}$

$=\displaystyle\frac{\cos \text{A}}{1+\sin \text{A}}\times{\displaystyle\frac{1-\sin \text{A}}{1-\sin \text{A}}}+\displaystyle\frac{1 + \sin \text{A}}{\cos \text{A}}$

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

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

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

$=\displaystyle\frac{2}{\cos \text{A}}$

$=2\sec \text{A}$.

**Proved.**

We have identified the pattern of $(1+\sin \text{A})$ in the first denominator as well as $\cos \text{A}$ in the numerator and have decided upon the suitable method to adopt to make the two denominators equal. The driving force were the primary objective to combine the two LHS terms, and then the secondary objective of making the denominators equal in as simple steps as possible.