## Mechanical reliance on routine procedures in solving problems

Often we find at high school level, math problems solved following a routine series of steps and in only one predictable way. The students learn to believe that generally problems are to be solved in only one mechanical routine way. Especially this we find in case of problems of the type,

Prove that, "Some expression " = "Some other expression".

In school math terminology, the "Some expression" is called **LHS** (short form of Left Hand Side) and the "Some other expression" as **RHS **(short form of Right Hand Side). This type of problems **occur ****abundantly **in Elementary Trigonometry of proving Identities.

These solutions use a **conventional approach ***of going towards the solution from LHS to RHS* (or initial state to goal state) usually through many steps using the **expansion of the LHS expression** and then **simplification or consolidation of the numerous expanded terms** towards the form of expression on the right hand side, that is, the RHS.

The

in this type of of conventional solutions is themost important element absentand theelement of Analysis,reasons why a particular path of solution is followed.

This approach has** two important disadvantages**,

- Not only does this approach
**take considerable amount of time and effort**, but*because of large number of steps*,**chances of error is much higher**in this approach. - This mechanical approach
*relies heavily on manipulation of terms*. In fact, if students follow only this approach of solving problems, they may tend to become used to mechanical and procedural thinking**using low level mathematical constructs without using the problem solving**abilities of the student**suppressing their inherent creative and innovative out-of-the-box thinking abilities.**

While solving the problem example we have chosen this time, our objectives will be two and both important. We will,

- show and encourage you to solve a problem in many ways. This is
, one of the most powerful**practice of Many Ways Technique**, and,**problem solving skill improving techniques** - show an example of problem solver's reasoning in devising
This is application of the**an unusual method of solution that uses only the simplest of the relevant formulas.**that is crucial to expand your general problem solving skillsets or abilities.**Less facts more procedures approach**

### Problem example

**Prove the identity:**

$\displaystyle\frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \frac{1 + \sin \theta }{\cos \theta}$

**First try to solve this problem yourself and then only go ahead. **

**Try to solve the problem in as many different ways as you can and judge and compare the advantages and disadvantages of the different methods of solution.**

**There is no better method than learning by yourself.**

### Conventional solution 1

We will see the first possible way this problem may be solved conventionally.

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

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

$=\displaystyle\frac{(\sec \theta + \tan \theta) - (\sec^2\theta - \tan^2\theta)}{\tan \theta - \sec \theta + 1}$, as $\sec^2\theta - \tan^2\theta = 1$

$=\displaystyle\frac{(\sec \theta + \tan \theta)(1 - sec \theta + \tan \theta)}{\tan \theta - \sec \theta + 1}$

$=\displaystyle\frac{(\sec \theta + \tan \theta)(\tan \theta - \sec \theta + 1)}{\tan \theta - \sec \theta + 1}$

$=\sec \theta + \tan \theta$

$=\displaystyle\frac{1}{\cos \theta} + \displaystyle\frac{\sin \theta}{\cos \theta}$

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

### Conventional solution 2 using rich trigonometric concept

The second possible way this problem may be solved conventionally,

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

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

$=\displaystyle\frac{\displaystyle\frac{1}{\sec \theta - \tan \theta} - 1}{\tan \theta - \sec \theta + 1}$, as $\sec^2\theta - \tan^2\theta = 1$, or, $\sec \theta + \tan \theta = \displaystyle\frac{1}{\sec \theta - \tan \theta}$, this we call a rich Trigonometric concept derived from basic concepts

$=\displaystyle\frac{1 - \sec \theta + \tan \theta}{(\sec \theta - \tan \theta)(\tan \theta - \sec \theta + 1)}$

$=\displaystyle\frac{\tan \theta - \sec \theta + 1}{(sec\theta - tan\theta)(\tan \theta - \sec \theta + 1)}$

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

$=\sec \theta + \tan \theta$, applying the rich concept formula a second time

$=\displaystyle\frac{1}{\cos \theta} + \displaystyle\frac{\sin \theta}{\cos \theta}$

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

### A new way to the solution in a few steps

As a part of efficient problem solving process, **the very first step** that you must take **is to analyze the problem.**

*In any problem solving, math or otherwise, this must be the first step.* You must start with analyzing the problem statement.

Without the first step of Problem analysis, no efficient problem solving is possible.

A corollary,

In competitive exams, and also in competitive work environment, the first step of problem analysis is crucial for success.

*The better and quicker you are able to analyze a problem, the faster you would reach the desired solution.*

#### Problem analysis

**The first step** in analyzing this type of problem is to see * how close are the goal state and the initial state.* In this case

*we conclude,*

**as there are no similarities between the RHS and the LHS at all,**

Inherently the given expressionhas the simplification resources in itselfwhich if used will lead us automatically to the simple goal state quickly.

**Secondly** we find the goal state RHS is in terms of $sin$ and $cos$ functions while the given expression is in one level higher $tan$ and $sec$ functions. So we take the clear decision to transform the given expression in terms of $sin$ and $cos$ in one single step. This should make further simplification easier. In trigonometry we can encapsualte this concept as a general technique, the **Goal form matching technique,**

Transform the level of the input or given expression to the level of the goal or target expression first.

Based on the analysis at the first stage and after applying goal form matching technique, now we take the bold step of * multiplying and dividing the given expression by the target expression.* The

*for this unusual action is:*

**reasoning**As the target expression has no similarity with the given expression, when we multiply and divide the given expression by the target expression, especially as both the given and the target expressions are fractional terms, the multiplied part will remain untouched while the

, thusdivided part has to resolve itself to unityleaving the multiplied part as the RHS or the solution directly.

**Aside: Psychology and process of problem solving by End State Analysis and deductive reasoning:** The desired goal to reach undoubtedly rank highest in importance in your mind among all other information about the problem as your natural tendency is to reach the goal state in quickest possible time.

*This pre-eminence of importance of the desired end state or goal state focuses your attention naturally on this end state when you know it. This is the case of proving identities..*

* What would you look for in the end state?*

*If it is a journey from one city to another, you study the distance to the destination from your starting point. You try to judge what kind of transportation along which path would take you to the destination in shortest possible time, isn't it? We assume here the importance of optimal journey, which is the case of any important problem solving.*

*The same happens in this case. You judge the end state (or RHS expression) with respect to the initial given state (or LHS expression). If somehow you find significant similarities between the two, it would be easy for you to span the gap between the two states quickly.*

*In majority of cases though there would be significant dissimilarity between the initial starting point and the desired end point. The similarity if at all there, would be hidden from casual inspection.*

*This is where the ability of key information discovery plays its prime role in solving the problem.*

*More often than not, ability to recognize useful common pattern, even if hidden, results in key information discovery.*

*If you don't know the desired goal state, from initial problem analysis you have to form possible desired goal states.*

*This is application of one of the most powerful problem solving resources that we are aware of - the End State Analysis. If you want to know more you can refer to it here.*

*Though usually there will be inherent similarities between the goal state and the initial state, in some cases there will be total dissimilarity. In such cases, the dissimilarity itself will be a key information for deciding the efficient course of action in solving the problem.*

**Problem solver's solution**:

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

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

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

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

$\hspace{35mm}\times{\displaystyle\frac{1+ \sin \theta}{\cos \theta}}$

$=\displaystyle\frac{\sin \theta - \cos \theta + 1}{(\sin \theta - 1) + \cos \theta}\times{\frac{\cos \theta}{(\sin \theta + 1)}}$

$\hspace{35mm}\times{\displaystyle\frac{1+ \sin \theta}{\cos \theta}}$

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

$\hspace{35mm}\times{\displaystyle\frac{1+ \sin \theta}{\cos \theta}}$

$=\displaystyle\frac{\sin \theta \cos\theta - \cos^2\theta + \cos \theta}{\sin \theta \cos \theta - \cos^2\theta + \cos \theta }$

$\hspace{35mm}\times{\displaystyle\frac{1+ \sin \theta}{\cos \theta}}$

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

This is a method that directly reaches the solution breaking down the barrier in a way that is not usual, but yes it reaches the destination with assurance and * in a few steps* notwithstanding the seemingly involved deduction in between.

### Less facts more procedures approach

You would notice that in this solution we have used the simplest level of concepts or formulas in Trigonometry along with problem analysis and deductive reasoning. If in general while you solve a problem in any situation, you rely more on the simplest concepts along with your problem solving techniques and strategies that are more general with wide applicability, finally you would find that you need to remember least amount of basic concepts and a few powerful problem solving techniques or procedures in solving a large variety of problems.

Practice and ability to use this approach enables you to quickly arrive at the most elegant solution with least memory load and least effort.

**End note:** There is at least one more solution to this problem. If you can, find it out. More importantly, our recommendation as always, would be to evaluate the solutions yourself and choose the one that suits you best. Be a learner and judge yourself

**And Always think:** is there any other shorter better way to the solution?

### Resources on Trigonometry and related topics

You may refer to our useful resources on Trigonometry and other related topics especially algebra.

#### Tutorials on Trigonometry

**Basic and rich concepts in Trigonometry and its applications**

**Basic and Rich Concepts in Trigonometry part 2, proof of compound angle functions**

**Trigonometry concepts part 3, maxima (or minima) of Trigonometric expressions**

#### Efficient problem solving in Trigonometry

**How to solve not so difficult SSC CGL level problem in a few light steps, Trigonometry 9**

**How to solve a difficult SSC CGL level problem in a few concepual steps, Trigonometry 8 **

**How to solve not so difficult SSC CGL level problem in a few light steps, Trigonometry 7**

**How to solve a difficult SSC CGL level problem in few quick steps, Trigonometry 6**

**How to solve a School Math problem in a few direct steps, Trigonometry 5**

**How to solve difficult SSC CGL level School math problems in a few quick steps, Trigonometry 5**

**How to solve School Math problem in a few steps and in Many Ways, Trigonometry 4**

**How to solve a School Math problem in a few simple steps, Trigonometry 3**

**How to solve difficult SSC CGL level School math problems in a few simple steps, Trigonometry 4**

**How to solve difficult SSC CGL level 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 simple steps, Trigonometry 1**

**A note on usability:** The *Efficient math problem solving* sessions on **School maths** are **equally usable for SSC CGL aspirants**, as firstly, the "Prove the identity" problems can easily be converted to a MCQ type question, and secondly, the **same set of problem solving reasoning and techniques have been used for any efficient Trigonometry problem solving.**

On the other hand, * any SSC CGL competitive test MCQ problem can also be converted to a School Math problem suitably.* That's why

**the resources on Trigonometry should be useful for School students as well as SSC CGL test aspirants.**#### 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**

#### SSC CGL Tier II level question and solution sets on Trigonometry

**SSC CGL Tier II level Solution set 12 Trigonometry 3, questions with solutions**

**SSC CGL Tier II level Question set 12 Trigonometry 3, questions with answers**

**SSC CGL Tier II level Solution set 11 Trigonometry 2**

**SSC CGL Tier II level Question set 11 Trigonometry 2**

**SSC CGL Tier II level Solution Set 7 on Trigonometry 1**

**SSC CGL Tier II level Question Set 7 on Trigonometry 1 **

#### SSC CGL level question and solution sets on Trigonometry

**SSC CGL level Solution Set 77 on Trigonometry 7**

**SSC CGL level Question Set 77 on Trigonometry 7**

**SSC CGL level Solution Set 65 on Trigonometry 6**

**SSC CGL level Question Set 65 on Trigonometry 6**

**SSC CGL level Solution Set 56 on Trigonometry 5**

**SSC CGL level Question Set 56 on Trigonometry 5**

**SSC CGL level Solution Set 40 on Trigonometry 4**

**SSC CGL level Question Set 40 on Trigonometry 4**

**SSC CGL level Solution Set 19 on Trigonometry**

**SSC CGL level Question set 19 on Trigonometry**

**SSC CGL level Solution Set 16 on Trigonometry**

**SSC CGL level Question Set 16 on Trigonometry**

**SSC CGL level Question Set 2 on Trigonometry**

**SSC CGL level Solution Set 2 on Trigonometry**

#### Algebraic concepts

**Basic and rich Algebraic concepts for elegant solutions of SSC CGL problems**

**More rich algebraic concepts and techniques for elega****n****t solutions of SSC CGL problems**