**It didn't seem to be a particularly difficult problem with conventional solution visible not far away**

For some problems that don't look particularly difficult, the conventional steps to the solution may not seem to take long, but invariably it will be a mundane and wasteful few extra steps. The elegant lighted up way would have been just a few quick steps away from the goal, waiting to be believed and discovered, always.

In this session we will showcase a seemingly innocuous problem that doesn't seem to be difficult. With such problems oftentimes we take the conventional path. The first solution presented will thus be the deductive conventional one.

This is how we think and act, conventionally, most of the times.

Though this conventional solution does not take many steps, it involves squaring and simplification, an additional burden.

We will present two more solutions with gradually decreasing efficiency.

Contrasting these three, the fourth solution will show how quickly and effortlessly the solution can be reached **using the rich potential embedded in the nature of the problem itself**.

It is *not only about saving of time and cost*, **it is also about elegance and effortlessness.**

Before going ahead you should refer to our concept tutorials on Trigonometry,

**Basic and Rich Trigonometry concepts and applications**

**Basic and Rich Trigonometry concepts part 2, Compound angle functions**

*Trigonometry concepts part 3, maxima and minima of trigonometric expressions.*

**Chosen Problem.**

If $cot \theta + cosec \theta =3$, and $\theta$ an acute angle, the value of $cos \theta$ is,

- $1$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{3}{4}$

**First solution: Conventional approach with no early function conversion but involving quadratic expression with one step conversion to linear**

The problem looks simple at first glance and that created a comfort zone in mind. Knowing that $cosec^2 \theta - cot^2 \theta=1$, a feasible path to the solution seems to be, taking the $cot \theta$ on the RHS and squaring the equation so that only $cot \theta$ is left in the expression. From the value of $cot \theta$ we would get $tan \theta$ and then $cos \theta$ using again the relation $1+tan^2 \theta=sec^2 \theta$. Solution is visible not too far away.

The given expression is,

$cot \theta + cosec \theta =3$,

Or, $cosec \theta = 3 - cot \theta$.

Squaring,

$cosec^2 \theta=9-6cot \theta + cot^2 \theta$,

Or, $9-6cot \theta=cosec^2 \theta - cot^2 \theta=1$,

Or, $6cot \theta = 8$,

Or, $tan \theta = \displaystyle\frac{3}{4}$.

Squaring and adding 1,

$tan^2 \theta + 1=sec^2 \theta =\displaystyle\frac{25}{16}$,

Or, $sec \theta = \displaystyle\frac{5}{4}$, as $\theta$ is acute $sec \theta$ is not negative.

So,

$cos \theta=\displaystyle\frac{4}{5}$.

**Answer:** Option c: $\displaystyle\frac{4}{5}$.

We will now present a second possible solution.

#### Second solution: Early conversion of trigonometric functions and use of quadratic expression with two terms and single function variable

The decision taken in this solution is to convert both the input functions into $cos$ and $sin$ terms. The reason behind this decision must be that the target is a $cos$ function.

Given expression,

$cot \theta + cosec \theta =3$,

Or, $\displaystyle\frac{cos \theta}{sin \theta}+\displaystyle\frac{1}{sin \theta}=3$,

Or, $cos \theta=3sin \theta -1$,

Squaring,

$cos^2 \theta=9sin^2\theta - 6sin\theta +1$,

Or, $10sin^2\theta=6sin\theta$, we could reduce number of variables to one and terms to two,

Or, $sin \theta = \displaystyle\frac{3}{5}$, with $\theta$ acute $sin \theta \neq 0$

Or, $cos \theta = \sqrt{1-\displaystyle\frac{3^2}{5^2}}=\displaystyle\frac{4}{5}$.

**Answer:** Option c: $\displaystyle\frac{4}{5}$.

You might think at this point that no more variation in solution approach is possible. The following yet another solution is a strong example of **Many ways technique.**

#### Third solution: Early trigonometric function conversion involving quadratic expressions with three terms

In the previous solution you will find that we could adhere to the powerful* principle of minimum number of terms and variables* during the simplification process. In the solution we could reduce the number of terms to two in the quadratic expression of a single variable. This allowed us avoiding the additional

*burden of*

**solving a quadratic equation.**The * principle of minimum number of terms and variables* says,

If you can reduce the number of terms as well as the number of variables in the expression involved in simplification to the minimum, faster will you reach the solution.

The intial approach in this third solution is same as in previous solution. The $cot$ and $sec$ functions are converted to $cos$ and $sin$ functions and resulting expression squared but with an important difference.

Given expression,

$cot \theta + cosec \theta =3$,

Or, $\displaystyle\frac{cos \theta}{sin \theta}+\displaystyle\frac{1}{sin \theta}=3$,

Or, $cos \theta+1=3sin \theta$.

Squaring,

$cos^2 \theta+2cos \theta+1=9sin^2\theta$,

Or, $cos^2 \theta+2cos \theta+1=9(1-cos^2\theta)$,

Or, $10cos^2 \theta+2cos \theta -8=0$,

Or, $5cos^2 \theta +cos \theta -4=0$,

Or, $(5cos \theta - 4)(cos \theta +1)=0$,

As $\theta$ is acute, $cos \theta \neq -1$.

So,

$cos \theta =\displaystyle\frac{4}{5}$.

**Answer:** Option c: $\displaystyle\frac{4}{5}$.

Compare the shortcomings of the three solutions and identify the reasons behind the shortcomings.

At the last we will now present the elegant most efficient problem solver's solution.

#### Fourth solution: Elegant solution: Problem solver's approach using principle of friendly trigonometric function pair and linear expressions

We are already aware of the * principle of friendly trigonometric function pairs*. The results from this rich principle are simple and based on fundamental concepts but have far reaching potential in simplifying trigonometric expressions quickly.

Two of the results of this principle are,

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

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

In this problem, the given expression directly conforms to this principle and rest of the steps are trivial algebraic simplification in linear expressions.

Given expression,

$cot \theta + cosec \theta =3$,

Or, $\displaystyle\frac{1}{cosec \theta - cot \theta}=3$, only the LHS changes,

Or, $cosec \theta - cot \theta =\displaystyle\frac{1}{3}$.

Adding the two equations,

$2cosec \theta=3+\displaystyle\frac{1}{3}=\displaystyle\frac{10}{3}$, $cot \theta$ is eliminated and we get the most desired single variable value

Or $cosec \theta = \displaystyle\frac{5}{3}$,

Or, $sin \theta =\displaystyle\frac{3}{5}$,

Or, $cos \theta =\sqrt{1-\displaystyle\frac{3^2}{5^2}}=\displaystyle\frac{4}{5}$, as $\theta$ is acute $cos \theta$ can't be negative.

**Answer:** c: $\displaystyle\frac{4}{5}$.

**Key concepts used:** * Key pattern identification* --

*--*

**rich concept of friendly trigonometric function pairs***--*

**basic algebraic concepts***--*

**solving simple linear equations**

**Many ways technique.**In this elegant solution, we have reduced the number of variables (in this case distinct trigonometric functions) to just one using simple algebraic manipulations and no squaring and dealing with quadratic expressions.

This refers to yet another basic principle of algebraic simplification, the * minimum order simplest solution principle.* Primarily, the more you increase the order (or power) of the terms in the deductive process, the more you deviate from the shortest path solution and make the problem harder to solve.

This fundamental algebraic principle of * minimum order simplest solution* states,

In a solution process, if you keep the order of the variables to the minimum, generally linear of unit power, you will have the simplest solution.

Adhering to this principle we always try to keep the expressions involved linear. You will notice that in the elegant solution also the expression were all linear.

**Important**

The elegant solution showcased above exemplifies,

- We should always start solving a problem with a strategy suitable and specific for the problem type and problem solving objectives. First objective is of course to solve the problem, but more importantly, solve the problem along the most efficient shortest path.
- Even if the conventional solution may seem to be within reach, one must analyze and search for a more efficient concept based solution before embarking on the conventional path.
- Any random approach will invariably take you along a longer and generally confusing path to the solution.
- The tendency and capacity to use basic and rich concepts for elegant and efficient solutions are not acquired in a day, these grow over a period of time and become reflexive in nature by constant search and execute of elegant solution concept structures.
- Specifically, the
have far reaching potential in simplifying many trigonometric expressions quickly. If in a problem such a friendly pair is explicitly present in an expression, it will invariably result in a quick solution.**friendly trigonometric function pair concepts** - To achieve elegant solutions in Trigonometry consistently,
**role of basic and rich algebraic principles is critically important.**

While solving the problem in four different ways, we have applied the problem solving skill improvement * Many ways technique* that tested our skill in solving a problem in multiple ways as well as gave us the opportunity to compare the multiple solutions with each other.

Even simple things need to be made simpler. Only then more difficult things can be made simple with ease.

### 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**

#### General guidelines for success in SSC CGL

**7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests**

#### 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.

#### 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 Trigonometry 1**

**SSC CGL Tier II level Question set 7 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**