## Efficient Problem solving techniques enable reaching goal in a few direct steps without goal manipulation

After crossing the first barrier in the process of proving a Trigonometric identity, if you go straight ahead without looking at the goal state on the RHS, you might reach a situation where you would have to start again from the goal state expression and backtrack to the simple result you had reached on the first go, not only an awkward solution but also technically an invalid one.

Instead, awareness of the similarities of the goal state with the problem state after every step taken, and use of the right kind of problem solving techniques will invariably take you to the goal state directly along the shortest path without using the goal state expression and backtracking.

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, or,*of manipulation of the RHS and LHS together to prove the identity, which should not be a valid way to the solve such a problem.*

Around us we find both forms of improper and complex problem solving without using the **most important elements** of problem solving, namely,

**problem analysis,**- use of powerful
**problem solving strategies for efficient solution, and** - the third very important element of
for understanding**comparing more than one path to the solution**and the others are not. This is the use of powerful**why a particular solution is the efficient solution****problem solving skill building Many ways technique.**

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.

These approaches have** two important disadvantages**,

- Not only does such an approach
**take considerable amount of time and effort**, but*because of large number of steps*,**chances of error is much higher**in the approach. - These mechanical approaches
*rely 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 three and all three important*. We will,

- Show and encourage you to
. This is**solve a problem in many ways**, one of the most powerful**practice of Many Ways Technique**.**problem solving skill improving techniques** - Show an example of problem solver's reasoning in
*using powerful trigonometric and general problem solving techniques to reach the solution in a straight and simple path***.** - On the way to the solution,
, that is, the RHS in any way,**we will avoid manipulating the goal expression****which we classify as an invalid method of solution.**

### Why manipulating the goal expression of RHS is an invalid method of solution

In a "**prove the identity problem",** the goal is given in the problem as the RHS. The problem solver may certainly use the * information contained in the RHS expression *in

*the efficient*

**determining***, just as we use the choice values in an equivalent MCQ problem. But as soon as the goal RHS expression is manipulated together with the main problem element in LHS expression, the fundamental*

**problem solving strategy**

**goal of solving the problem is significantly compromised.**The * basic goal in solving any problem*, be it in school or in competitive exams, can only be,

To test the

in the specific domain of the problem.problem solving ability of the problem solver

**Note:** In real life problem solving, you can't use the goal solution, you can at best form a hypothesis of the goal state. Thus **manipulating the goal state in real life problem solving is an impossibility.**

Before going ahead, you may like to refer to the session on **Basic and rich Trigonometry concepts and applications. **

### Problem example

**Prove the identity:**

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

**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.**

### Efficient solution

We will see the first how this problem may be solved in a few simple steps without manipulating the RHS expression in any way.

$LHS=\displaystyle\frac{1}{cosec \theta - cot \theta} - \displaystyle\frac{1}{sin \theta}$

$=\displaystyle\frac{(cosec \theta + cot \theta)}{(cosec \theta + cot \theta)(cosec \theta - cot \theta)} - \displaystyle\frac{1}{sin \theta}$.

Here we are about the use the * Friendly function pair concept* of,

$\displaystyle\frac{1}{cosec \theta - cot \theta}$

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

$=\displaystyle\frac{cosec \theta + cot \theta}{cosec^2 \theta - cot^2 \theta}$

$=cosec \theta + cot \theta$, as $cosec^2 \theta - cot^2 \theta=1$.

Thus,

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

and vice versa and thus we call the two functions $cosec \theta$ and $cot \theta$ as one of the Friendly trigonometric function pairs.

So in our problem,

$LHS=\displaystyle\frac{(cosec \theta + cot \theta)}{(cosec \theta + cot \theta)(cosec \theta - cot \theta)} - \displaystyle\frac{1}{sin \theta}$

$=\displaystyle\frac{cosec \theta + cot \theta}{cosec^2 \theta - cot^2 \theta}- \displaystyle\frac{1}{sin \theta}$

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

We look at the goal expression on the RHS at this point, and find that we have nearly formed the first term, as well as the the second term of the goal expression. This is use of* End state analysis approach* where we compare the present problem state with the end goal state to find how similar those are in form.

Now we need just a bit of transformation and regrouping of terms. These techniques of * term transformation* and

*are frequently used in efficient problem solving.*

**regrouping of terms**So,

$LHS=cosec \theta + cot \theta -\displaystyle\frac{1}{sin \theta}$

$=\displaystyle\frac{1}{sin \theta} - (cosec \theta - cot \theta)$.

At this point we can directly substitute $\displaystyle\frac{1}{cosec \theta + cot \theta}$ for $(cosec \theta - cot \theta)$. Instead again proceeding formally we multiply both numerator and denominator of the second term by $(cosec \theta + cot \theta)$,

$LHS=\displaystyle\frac{1}{sin \theta} - (cosec \theta - cot \theta)$

$=\displaystyle\frac{1}{sin \theta} - \displaystyle\frac{(cosec \theta + cot \theta)(cosec \theta - cot \theta)}{(cosec \theta + cot \theta)}$

$=\displaystyle\frac{1}{sin \theta} - \displaystyle\frac{cosec^2 \theta - cot^2 \theta}{cosec \theta + cot \theta}$

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

Or, $LHS=RHS$. Identity proved.

We have shown every little expression manipulation, however obvious and trivial that may be, accompanied by the explanation why a step is being taken.

**Important concepts used:** **Friendly trigonometric function pair concept -- ****End state analysis approach -- Term transformation tehnique -- ****Regrouping of terms technique.**

### Conventional solution using manipulation of goal RHS expression

The second possible way this problem may be solved conventionally,

$LHS=\displaystyle\frac{1}{cosec \theta - cot \theta} - \displaystyle\frac{1}{sin \theta}$

$=\displaystyle\frac{(cosec \theta + cot \theta)}{(cosec \theta + cot \theta)(cosec \theta - cot \theta)} - \displaystyle\frac{1}{sin \theta}$

$=\displaystyle\frac{cosec \theta + cot \theta}{cosec^2 \theta - cot^2 \theta}- \displaystyle\frac{1}{sin \theta}$

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

$=cosec \theta +cot \theta -cosec \theta$

$=cot \theta$

In this possible conventional solution, we have used the * resource of friendly function pair* alright but

*or the term transformation and regrouping of terms technique, the conventional process goes on to the*

**instead of using the powerful end state analysis approach**

**apparently attractive goal of simplifying the result to its simplest form.**But as soon as we do that * we lose the opportunity* to move on to the goal state in a few more simple steps

*and*

**using term transformation***directly,*

**term regrouping***at all.*

**without manipulation of goal state RHS**We are * left with no other alternative than to start manipulating and simplying the RHS* to the result that we just got.

$RHS=\displaystyle\frac{1}{sin \theta} - \displaystyle\frac{1}{cosec \theta + cot \theta}$

$=\displaystyle\frac{1}{sin \theta}-\displaystyle\frac{(cosec \theta - cot \theta)}{(cosec \theta - cot \theta)(cosec \theta + cot \theta)}$

$=\displaystyle\frac{1}{sin \theta}-\displaystyle\frac{cosec \theta - cot \theta}{cosec^2 \theta - cot^2 \theta}$

$=cosec \theta - (cosec \theta - cot \theta)$

$=cosec \theta +cot \theta -cosec \theta$

$=cot \theta$

$=LHS$.

#### Summary of Problem analysis and Efficient problem solving steps

**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 find goal state and problem state both have one common expression of $\displaystyle\frac{1}{sin \theta}$, but in opposite signs. As we don't find any further similarity

*of the problem and immediately identify it to be a part of a*

**we examine the first term**

**friendly trigonometric function pair.**Now knowing that this term can be transformed easily into $cosec \theta + cot \theta$, we look at the goal RHS expression. The * goal form matching technique* comprising of

*and*

**term transformation***to reach the exact goal expression of the RHS becomes transparently clear at this stage itself.*

**regrouping of terms**Rest is straightforward execution in a few simple steps **avoiding any form of manipulation of the goal RHS expression.**

**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 preeminence 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.*

**End note:** If you can, try to find more number of different solutions to the problem. 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 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 in 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**