## Reliance on mechanical procedures in solving problems

Often we find at high school level, math problems solved in a long series of steps. 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 long solutions use a **conventional approach ***of going towards the solution from LHS to RHS* (or initial state to goal state) 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.

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 are 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.**

Let us go through the process of solving a Trigonometric Identity problem **to appreciate the difference between the conventional approach and the problem solver's intelligent approach.** The *difference is always significant and measurable.*

### Problem example

**Prove the identity:**

\begin{align} (cos\theta & – cosec\theta)^2 + (sin\theta – sec\theta)^2 \\ & = (1 – sec\theta{cosec\theta})^2 \end{align}

**First try to solve this problem yourself and then only go ahead. It may very well be possible that you would be able to solve it quickly in a few steps.**

### Conventional solution

Usually a conventional approach in this case will involve a long deduction process as all the terms are in the form of squares of expressions.

The square expressions are expanded, then **accumulated together suitably** so that further simplification is possible.

After accumulating together the friendly terms, the expressions are simplified and then again retransformed back to the desired expression of the RHS.

This is **an abstract description of the generally followed conventional problem solving**. Let us see actually how it is done in our problem example case.

Taking the Left hand side expression $(cos\theta – cosec\theta)^2 + (sin\theta – sec\theta)^2$ and expanding we get LHS as,

\begin{align} & (cos\theta – cosec\theta)^2 + (sin\theta – sec\theta)^2 \\ & = (cos^2\theta - 2cos\theta{cosec\theta} + cosec^2\theta) \\ &\hspace{10mm} + (sin^2\theta - 2sin\theta{sec\theta} + sec^2\theta) \\ & = (sin^2\theta + cos^2\theta) \\ &\hspace{10mm} - 2\left(cos\theta{cosec\theta} + sin\theta{sec\theta}\right) \\ & \hspace{10mm} + (cosec^2\theta + sec^2\theta) \\ & = 1 - 2\left(\displaystyle\frac{cos\theta}{sin\theta} + \displaystyle\frac{sin\theta}{cos\theta}\right) \\ & \hspace{10mm} + \left(\displaystyle\frac{1}{sin^2\theta} + \displaystyle\frac{1}{cos^2\theta}\right) \\ & = 1 - 2\left(\displaystyle\frac{cos^2\theta + sin^2\theta}{sin\theta{cos\theta}}\right) \\ & \hspace{10mm} + \left(\displaystyle\frac{cos^2\theta + sin^2\theta}{sin^2\theta{cos^2\theta}}\right) \\ & = 1 - 2\left(\displaystyle\frac{1}{sin\theta{cos\theta}}\right) \\ & \hspace{10mm} + \left(\displaystyle\frac{1}{sin^2\theta{cos^2\theta}}\right) \\ & = 1 - 2sec\theta{cosec\theta} + (sec\theta{cosec\theta})^2 \\ & = (1 - sec\theta{cosec\theta})^2 \\ & = RHS \end{align}

### Efficient solution in a few steps

Instead of this conventional approach **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 analyze a problem, the faster you would reach the desired solution.*

#### Problem analysis

As **you have already the goal state in the form of the expression on the RHS** of the $"="$ symbol, your immediate reaction would be to examine **how much similarity does the RHS expression have compared to the expression you would start with on the LHS**.

**Aside: Psychology and process of problem solving by End State Analysis:** 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 all cases though there would be significant dissimilarity between the initial starting point and the desired end point. The similarity would invariably be there but it 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. *

In most cases this similarity would certainly be there (as you can certainly transform the LHS into RHS, you may not know how, but you should most possibly find a crucial similarity between the two). But this similarity always will be hidden behind a barrier. Your job is to look through the barrier to discover the key information.

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

We would strongly **recommend you to follow this path**. Reason is - *we want you to find the key information for quick solution before you go through our explanation*. **If you solve a problem elegantly yourself your learning will be maximum.**

Take a pause, go through * the solution using End State Analysis*.

**Learn what it is**and

**how to use the concept**and

**solve this problem in a few steps yourself.**

#### Key information discovery

Our goal is transforming each of the two terms in the LHS to the form of expression in the RHS. Let us take the first term $(cos\theta - cosec\theta)^2$.

Our obvious attention goes to the first term of the expression and we are clear that to transform this term of $cos\theta$ to 1, the value of this first term in target end state expression, we must factor $cos\theta$ out of the brackets.

In doing so, we need to multiply the second term by inverse of $cos\theta$. That we do. But when the $cos\theta$ is factored out of the brackets it comes out squared. Thus for the first term we get,

\begin{align} (cos\theta & - cosec\theta)^2 \\ & = cos^2\theta(1 - sec\theta{cosec\theta})^2 \end{align}

Immediately you expect a simlar convenient result to come out of the second expression. Indeed that is exactly what happens. For the second term we get,

\begin{align} (sin\theta & – sec\theta)^2 \\ & = sin^2\theta(1 - sec\theta{cosec\theta})^2 \end{align}

Now you know the solution to be only a step away. We have LHS as,

\begin{align} & (cos\theta – cosec\theta)^2 + (sin\theta – sec\theta)^2 \\ & = cos^2\theta (1 – sec\theta{cosec\theta})^2 \\ & \hspace{10mm} + sin^2\theta(1 – sec\theta{cosec\theta})^2 \\ & = (sin^2\theta + cos^2\theta) (1 – sec\theta{cosec\theta})^2 \\ & = (1 – sec\theta{cosec\theta})^2\end{align}

This is by far the quickest way to reach the solution in just three steps and is based fully on examining and using the similarity between the final state and the initial state (here the final state is given and is only one. You needed to find the way to reach the final state from the initial state).

Let us review our reasoning process more closely.

### Deductive reasoning

On comparison of the RHS and LHS we find both to be squares. We form our immediate conjecture,

It is highly likely that the RHS expression in exactly the same form is hidden in each of the two squared expressions on the LHS.

At this second stage now we shift our comparison of the RHS expression with each of the two terms on the LHS. We take up the first term.

At this third stage, the first term of RHS is 1 and the first term in our first expression under focus is $cos\theta$. Without any hesitation, our deductive reasoning mechanism dictates us to transform this $cos\theta$ to 1. How? Easy, just by factoring it out of the brackets. As soon as we do this, the puzzle is solved in our mind.

Rest is routine.

**End note:** This is not the only example of this type. You would find very large number of this and other type of problems where you would invariably be able to improve upon the time and effort in conventional solution.

**Our recommendation:** **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.**### SSC CGL question and solution sets on Trigonometry

**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 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 3**

**SSC CGL level Question set 19 on Trigonometry 3**

**SSC CGL level Solution Set 16 on Trigonometry 2**

**SSC CGL level Question Set 16 on Trigonometry 2**

**SSC CGL level Question Set 2 on Trigonometry 1**

**SSC CGL level Solution Set 2 on Trigonometry 1**

### 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**