## A few simple steps to solution compared to conventional complex procedure

Many times we find at high school level, math problems are solved following a long series of steps. This is what we call * conventional approach to solving problems*.

This approach not only involves usually **a large number of step**s, in most cases, the steps themselves introduce a higher level of **complexity**.

These two aspects of conventional problem solving together result in what we call * costly and inefficient problem solving*.

In school math or real life problems, this is the path commonly followed by most. In school math, among other topics we find this form of solution abundantly 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 **occurs ****aplenty **in Elementary Trigonometry of proving Identities.

* We have covered one such example* already, This is the second example.

These long and mostly complex 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 and increased complexity*,**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 for 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 another 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:**

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

*First try to solve this problem yourself and then only go ahead. You might be able to reach the elegant solution to this problem yourself.*

### Conventional solution

Usually a conventional approach in this case will involve a long and complex deduction process involving a large amount of mathematical manipulation.

We combine the two terms in a brute-force method (there can be other such conventional methods) to transform the LHS as,

\begin{align} & \displaystyle\frac{sin\theta(1 + cos\theta) + tan\theta(1 - cos\theta)}{(1 - cos\theta)(1 + cos\theta)} \\ & = \displaystyle\frac{sin\theta + sin\theta{cos\theta} + tan\theta - sin\theta}{1 - cos^2\theta} \\ & = \displaystyle\frac{sin\theta{cos\theta} + tan\theta}{sin^2\theta} \\ & = cosec^2\theta(sin\theta{cos\theta} + tan\theta) \\& = cosec\theta{cos\theta} + cosec\theta{sec\theta} \\ & = sec\theta{cosec\theta} + cot\theta\end{align}

In this method we went through a straightforward approach of combining the two fraction terms in the LHS, and going though carefully the simplification process relying on our knowledge of conversion of trigonometric terms so that we can move towards the end state of the given expression with certainty and no error.

As such there is nothing wrong with this conventional straightforward approach.

On second thoughts though, we find we have to go through a lot of trigonometric procedures and the complexity level of the deductions is a bit high.

In the discipline of * Efficient Math Problem Solving* we emphasize repeatedly, that

Always use your efficient problem solving strategies more and the mathematical procedures less in solving any math problem.

This is broadly our * Less facts and more procedures approach *that we would deal with in detail later. Nevertheless,

*each of the efficient math problem solving examples*actually use this more abstract problem solving approach. Formally we state this

*Less facts and more procedures approach*as,

For solving any problem, use least amount of unconnected facts and more of interlinked concepts and efficient problem solving procedures.

In maths problem solving, this approach boils down to the strong recommendation highlighted by many eminent mathematicians among others,

Do more of problem solving and less of maths.

### Efficient solution in a few steps

Instead of this conventional approach, **the very first step** that you must take as usual **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

As a first step of analysis in such math problems, we follow the* End State analysis approach *invariably and

**compare the desired end state expression with the given initial state expression for**detecting

**similarities or dissimilarities**.

**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 the desired end state. 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 from your starting point **to the destination*. You try to judge what kind of transportation along which possible 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. *

*We add one aspect here, while studying the similarity between the end state and the intial state, many times the significant dissimilarity also may help you to find the key to the problem solution. In such cases, the dissimilarities help us in transforming the given state in the most efficient way to reach the end state quickly.*

In this problem we observe, though the **terms of the** **given expression** are **fractions involving denominators**, the **end state goal expression on the RHS does not have any fractions**. This is the **significant dissimilarity** that would guide us to our main strategy to deal with problem elegantly.

#### Key information discovery

Thus we form our **goal as transforming each of the two terms in the LHS to eliminate the denominators** as quickly and as elegantly as possible.

This objective is further strengthened when we observe both the denominators in the form of $x + y$ or $x - y$.

This is a valuable information to us because of our knowledge of the frequently used * Technique of Rationalization* using the important concept,

$(x + y)(x - y) = x^2 - y^2$.

While rationalizing a fraction term (usually this technique is applied in rationalizing surd expressions involving terms in the denominator such as $\sqrt{2} + 1$), we multiply both the numerator and the denominator of the fraction with a suitable expression such that the denominator is highly simplified. Most often this simplification happens using the concept of $x^2 - y^2 = (x + y)(x - y)$.

In Trigonometry, rationalization of this form further uses the most fundamental concept,

$sin^2\theta + cos^2\theta = 1$,

Or, $1 - cos^2\theta = sin^2\theta$,

Or, $1 - sin^2\theta = cos^2\theta$.

Armed with this result of our analysis along with the concept of rationalization technique, we proceed to rationalize each term of the LHS to get,

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

$\hspace{25mm} + \displaystyle\frac{(1 - cos\theta)tan\theta}{1 - cos^2\theta}$

Or, $LHS = cosec\theta(1 + cos\theta) $

$ \hspace{25mm} + sec\theta{cosec\theta}(1 - cos\theta)$.

At this stage, we have used the concept, $1 - cos^2\theta = sin^2\theta = \displaystyle\frac{1}{cosec^2\theta}$.

Or, $LHS = cosec\theta + cot\theta $

$ \hspace{25mm} + sec\theta{cosec\theta} - cosec\theta$

$ \hspace{20mm} = RHS$.

This is by far the quickest way to reach the solution in just three steps and is based fully on examining and using the **dissimilarity between the final state and the initial state** and deciding that **rationalization would make the states **(or expressions)** more similar**.

**Note:** In this solution, we have skipped **a few trivial steps** that you may not be allowed to skip in your formal prescribed school procedure. Saving in that case won't be so much as in reduced number of steps, rather the saving would be in following a less complex path to the solution resulting in saving in time and reducing chances of error.

### Deductive reasoning

On comparison of the RHS and LHS we find the * given expression having terms with denominators* while the

*goal state expression is without any fraction term*involving denominators.

Furthermore, inspecting the denominators of the fraction tems, * we observe a pattern in each* that should enable us to easily

**eliminate the denominators by rationalization.**This is where we use our **Pattern recognition technique.**

Thus we form **our first conjecture** as,

The immediate task is to rationalize each term of the given expression using concepts $x^2 - y^2 = (x + y)(x - y)$ together with $1 - cos^2\theta = sin^2\theta$.

At the second stage, rationalizing and transformation of the given terms using simple trigonometric relations quickly take us to the desired goal expression.

**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 conceptual 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 82 on Trigonometry 8**

**SSC CGL level Question set 82 on Trigonometry 8**

**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**