## Problem solving approach gave us the solutions in a few steps

In this session, we will again take up two selected NCERT Trigonometry problems of class 10 standard which we will solve in a few steps.

In the process, we will **highlight the problem solving approach**, in which, by applying **useful pattern identification, time tested trigonometric and algebraic expression simplifying strategies and powerful problem solving techniques, ***quick solution of the problems could be achieved in a few steps.*

Approach is to always *analyze the problem for finding useful patterns and adopt applicable methods and strategies to solve the problems in mind.*

We will provide explanations along with solutions.

Let us solve the problems to show you what we mean by the above statements. We urge you to try to solve the problems yourself before going through the solutions.

### Solutions to NCERT class 10 level Trigonometry problems—Set 4

#### Problem 1.

Prove the identity,

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

#### Solution 1. Problem analysis, Primary objective analysis and first pattern identification

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, $\text{LHS} \Rightarrow \text{RHS}$.

On **initial problem analysis** deciding on the primary objective is obvious,

We must eliminate the denominator.

As such there is no direct way of doing that. But we observe the first key pattern,

All three terms of the numerator and denominator are same, except for the signs.

We make our **first conclusion**—eliminating the denominator will somehow depend on this first key pattern. We don't know how at this point but, we know, it must.

Can we use $(a+b)(a-b)=a^2-b^2$ on the numerator denominator? Briefly we looked into the possibility and mentally going through just one step rejected it.

Then we recalled a **general trigonometric expression simplifying strategy** that we have formed through our experience in solving many identities of complex forms. Let's state it,

In general, use of $(\sec \theta \pm \tan \theta)$ or $(\text{cosec } \theta \pm \text{cot } \theta)$ simplifies an expression involving $\sin \theta$ and $\cos \theta$ much faster and in fewer number of steps, than using the more well-known relation $(sin^2 \theta+cos^2 \theta=1)$.

The reason of supremacy of **$\sec$-$\tan$** and **$\text{cosec}$-$\text{cot}$** over **$\sin$-$\cos$** in solving more complex problems lies in the **inverse relationships** between their complementary additive subtractive expressions,

$\sec \theta-\tan \theta=\displaystyle\frac{1}{\sec \theta + \tan \theta}$, and

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

Because of the high simplifying potential in complex trigonometric problem solving, we consider these two pairs as * primary friendly trigonometric function pairs*. We include

**$\sin$-$\cos$**also in this category, but as a

*secondary pair of less potential.*

#### Solution 1: End state analysis confirmed our decision to transform in terms of $\text{cosec } \text{A}$ and $\text{cot } \text{A}$

Following * End state analysis approach*, a powerful general problem solving strategy, as we examined the RHS, presence of $(\text{cosec } \text{A}+\text{cot } \text{A})$ helped us make up our mind only to use this primary friendly trigonometric function pair for quick simplification.

We have to convert the LHS in terms of $\text{cosec } \text{A}$ and $\text{cot } \text{A}$.

As expected, *reaching the result from this point took less than 20 secs in only a few steps.*

Let us show you the deductive steps.

#### Solution 1: Problem solving execution: Deductive steps

Accordingly, converting the LHS in terms of, $\text{cot } \text{A}$ and $\text{cosec } \text{A}$,

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

$=\displaystyle\frac{\text{cot } \text{A}-1+\text{cosec } \text{A}}{\text{cot } \text{A}+1-\text{cosec } \text{A}}$, dividing numerator and denominator by $\sin \text{A}$,

$=\displaystyle\frac{\text{cosec } \text{A} + \text{cot } \text{A}-1}{1-(\text{cosec } \text{A}-\text{cot } \text{A})}$, rearranging the terms for ease of understanding,

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

$=(\text{cosec } \text{A}+\text{cot } \text{A})\times{\displaystyle\frac{\text{cosec } \text{A}+\text{cot } \text{A} - 1}{\text{cosec } \text{A}+\text{cot } \text{A} - 1}}$

$=\text{cosec } \text{A}+\text{cot } \text{A}$

Proved.

You may try other methods.

This is a problem where the solution in a few steps is not directly visible and you have to go deeper and use your experience and concept based trigonometric problem solving strategies.

We will show a second approach to reasoning that enables you to take the crucial decision to convert the LHS in terms of $\text{cosec } \text{A}$ and $\text{cot } \text{A}$ faster.

#### Reasoning by Working backwards approach

The thinking goes like this from the **end to the beginning**—

- If at the end of simplification, only the factor $(\text{cosec } \text{A}+\text{cot } \text{A})$ is left, any other factors present before this final stage must have been cancelled out leaving only this factor,
- To have the factor $(\text{cosec } \text{A}+\text{cot } \text{A})$ in the numerator, the most suitable way is to have its complementary expression, $(\text{cosec } \text{A}-\text{cot } \text{A})$ in the denominator, at the stage just before the previous stage above. Remember that we are moving backwards.
- To have the expression, $(\text{cosec } \text{A}-\text{cot } \text{A})$ in the denominator then, the given expression must be transformed to $\text{cosec } \text{A}$ and $\text{cot } \text{A}$,
- Finally reaching the beginning, we examine the given expression and find that, yes, this is the way to go.

This powerful problem solving approach of Working backwards can be considered as an offshoot of more general problem solving approach of End state analysis. If you are curious, you may go through our articles on the two problem solving approaches for *general purpose problem solving* including solving problems in real life,

* Solving tough problems elegantly using End state analysis approach*, and

* Problem solving by Working backwards approach*.

#### Problem 2

Prove the identity,

$\sqrt{\displaystyle\frac{1+\sin \text{A}}{1-\sin \text{A}}}=\sec \text{A}+ \tan \text{A}$

#### Solution 2: Problem analysis, primary objective analysis and key pattern identification

The primary objective is obvious,

We have to transform the terms or expressions under square root of LHS in the form of squares to make those free of the square root.

The key pattern that would achieve this objective is also clear,

Multiply the denominator and numerator by $(1+\sin \text{A})$.

The numerator, free of square root becomes then, $(1+\sin \text{A})$, and the denominator, $\cos \text{A}$.

Straightforward division of the numerator by the denominator results in, $\sec \text{A} + \tan \text{A}$, the RHS.

Proved.

Let us show the steps.

#### Solution 2: Problem solving steps

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

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

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

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

$=\sec \text{A}+\tan \text{A}$

Proved.

This is a less than twenty seconds problem solvable in mind.

#### End note

We consider the approach to the solution more important than the solution itself. It is a way of thinking, not just solution of a problem.

### Further reading materials on Trigonometry

#### NCERT solutions for class 10 maths

**NCERT solutions for class 10 maths Ttrigonometry Set 6**

**NCERT solutions for class 10 maths Trigonometry Set 5**

**NCERT solutions for class 10 maths Trigonometry Set 4**

**NCERT solutions for class 10 maths Trigonometry Set 3**

**NCERT solutions for class 10 maths Trigonometry Set 2**

**NCERT solutions for class 10 maths Trigonometry Set 1**

#### Class 10 Maths

**How to solve school math problems in a few direct steps Trigonometry 5**

**How to solve school math problems in a few steps and many ways Trigonometry 4**

**How to solve school math problems in a few simple steps Trigonometry 3**

**How to solve school math problems in a few simple steps Trigonometry 2**

**How to solve school math problems in a few steps Trigonometry 1**

#### Tutorials, question and answer sets for competitive exams valuable for school level

You may refer to a fair amount of * concise tutorials* and

*created for competitive exams at the page containing list of links,*

**MCQ type question and answer sets**Related articles on NCERT solutions follow.

### NCERT Solutions for Class 10 Maths

#### Real Numbers

**NCERT Solutions for Class 10 Maths on Real numbers part 1, Euclid’s division lemma puzzle solutions**

#### Introduction to Trigonometry

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 6**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 5**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 4**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 3**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 2**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 1**