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

It was an awkward problem, but application of a powerful general principle provided the breakthrough

How to solve SSC CGL level difficult problem in quick conceptual steps trigonometry 8

This problem in contrast to the earlier one looked difficult. The challenge was on - would we be able to achieve its elegant solution? Not always we know about the end in the beginning of our journey, but finally concept based strategic approach provided the initial breakthrough and then led us to the elegant solution in a few more steps.

In this session we will showcase a hard problem that seemed to have no elegant solution. And then, use of a more general problem solving principle quickly led us to familiar grounds. It was application of a series of rich powerful concepts that produced the elegant solution in a few steps, quickly.

We will present first the cumbersome conventional solution and then the elegant solution for critical comparison and appreciation of the power of problem solving principles and techniques in producing elegant and efficient solution.

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 $(a^2-b^2)sin \theta + 2abcos \theta=a^2+b^2$ then the value of $tan \theta$ is,

  1. $\displaystyle\frac{1}{2ab}(a^2+b^2)$
  2. $\displaystyle\frac{1}{2}(a^2-b^2)$
  3. $\displaystyle\frac{1}{2}(a^2+b^2)$
  4. $\displaystyle\frac{1}{2ab}(a^2-b^2)$

First solution: Conventional approach that involves raising the powers of functions and extensive deduction

As value of $tan \theta$ is wanted, dividing by $cos \theta$ the $sin \theta$ is converted to $tan \theta$ creating an additional $sec \theta$ in the process.

But $sec \theta$ and $tan \theta$ being a friendly trigonometric function pair having the relation, $sec^2 \theta=tan^2 \theta +1$ at the basic concept level, it is expected that solution can be reached by raising the power of the variable functions to 2 by squaring,

$(a^2-b^2)sin \theta + 2abcos \theta=a^2+b^2$,

Or, $(a^2-b^2)tan \theta + 2ab=(a^2+b^2)sec \theta$

Squaring the equation,

$(a^2-b^2)^2tan^2 \theta + 4abtan \theta(a^2-b^2) $

$\hspace{30mm}+ 4a^2b^2=(a^2+b^2)^2sec^2 \theta$,

Or, $(a^2-b^2)^2tan^2 \theta + 4abtan \theta(a^2-b^2) $

$\hspace{30mm}+ 4a^2b^2=(a^2-b^2)^2sec^2 \theta+4a^2b^2sec^2 \theta$,

Or, $4abtan \theta(a^2-b^2) + 4a^2b^2$

$\hspace{30mm}=(a^2-b^2)^2(sec^2 \theta-tan^2 \theta)+4a^2b^2sec^2 \theta$,

Or, $(a^2-b^2)^2+4a^2b^2(tan^2 \theta+1)$

$\hspace{30mm}-4abtan \theta(a^2-b^2) - 4a^2b^2=0$,

Or, $(a^2-b^2)^2-4abtan \theta(a^2-b^2) +4a^2b^2tan^2 \theta=0$,

Or, $\left[(a^2-b^2)-2abtan \theta\right]^2=0$,

So,

$(a^2-b^2)-2abtan \theta=0$

Or, $tan \theta=\displaystyle\frac{1}{2ab}(a^2-b^2)$.

Answer: Option d: $\displaystyle\frac{1}{2ab}(a^2-b^2)$.

This is hard, extensive deduction based, and not a very reliable solution path.

Second solution: Elegant solution: Stage 1: Problem solving approach: Problem analysis revealed lack of harmony in the expression

The expression seemed to be discordant with lack of harmony and too many terms.

No clear elegant solution path could be seen at first analysis.

Let us explain what we mean by discordance and lack of harmony.

Principle of harmony or dicordance in expressions

The variables are the functions here. The first important characteristic is the two functions involved, $sin \theta$ and $cos \theta$ are friendly trigonometric function pair, though in comparison, this pair of functions does not have as much potential in simplification of expressions as the other two friendly trigonometric function pairs, $sec \theta$, $tan \theta$ and $cosec \theta$, $cot \theta$.

Each of these latter two have the powerful property of mutually inverse expression relationship of the form, $sec \theta + tan \theta=\displaystyle\frac{1}{sec \theta - tan \theta}$ which in general carries high potential in simplifying complex expressions easily and elegantly.

Coming back to the discordant form of the given expression, the discordance and lack of harmony originate from the term association of the friendly functions $sin \theta$ and $cos \theta$. the function $sin \theta$ is associated with a factor $a^2-b^2$ whereas its pair partner $cos \theta$ is associated with a term $2ab$ which is very different in structure and form from $a^2-b^2$. If the factor of $cos \theta$ were, say, $a^2+b^2$ we could have accepted the expression as an expression in harmony.

The principle of harmony or discordance in expressions states,

The more is the harmony or less is the dicordance in an expression, chances of existence of an elegant conceptual solution in a few steps will be that much more.

How to increase the harmony in the expression - first stage

On a closer look, a possibility of introducing significant harmony (still part harmony, not whole) in the expression could be identified.

If both sides of the equation are divided by $cos \theta$, thus breaking the discordant association of $cos \theta$ with $2ab$, in a single step we transform the equation in terms of more preferred friendly trigonometric function pair of $sec \theta$ and $tan \theta$ and on top of it form term associations of the variables, $sec \theta$ and $tan \theta$ that are similar in structure and form, that is, $(a^2+b^2)$ and $(a^2-b^2)$,

$(a^2-b^2)sin \theta + 2abcos \theta=a^2+b^2$,

$(a^2-b^2)tan \theta + 2ab=(a^2+b^2)sec \theta$.

In one simple action, the expression is transformed to a much more balanced and harmonious form with lesser discordance.

How to increase the harmony in the expression - second stage: apply principle of collection of friendly terms to create mutully inverse expressions in friendly trigonometric function pairs

Knowing the power of the mutually inverse expressions, $sec \theta +tan \theta$ and $sec \theta-tan \theta$ of the friendly trigonometric function pair, $sec \theta$, and $tan \theta$, the immediate next step in increasing the harmony in the expression further is to apply the principle of collection of friendly terms and bring the terms involving $sec \theta$ and $tan \theta$ together.

This is a familiar situation and decisions are easy to take with end state clearly visible. But in the very beginning it was not so. The key brekthrough was achieved by increasing the harmony or balance in the expression.

From the previous stage,

$(a^2-b^2)tan \theta + 2ab=(a^2+b^2)sec \theta$,

Or, $2ab=a^2(sec \theta - tan \theta) + b^2(sec \theta + tan \theta)$.

Second solution: Elegant solution: Stage 2: Resolving the last trace of discordance in the expression: Forming mutually inverse coefficient factors

With RHS transformed into a promising form, we turn our attention to the last offending discordant factor term $ab$ in the LHS.

This term is in no way is similar to the coefficients $a^2$ or $b^2$, but if we divide the equation by $ab$, not only is it removed from the expression, but also a pair of mutually inverse coefficient factors, $\displaystyle\frac{a}{b}$ and $\displaystyle\frac{b}{a}$ is formed in the RHS. This situation is still more welcome as the other two factors of the two terms, $sec \theta - tan \theta$ and $sec \theta + tan \theta=\displaystyle\frac{1}{sec \theta-tan \theta}$ are already present as a second pair of mutually inverse expressions.

From the previous stage,

$2ab=a^2(sec \theta - tan \theta) + b^2(sec \theta + tan \theta)$,

Or $2=\displaystyle\frac{a}{b}(sec \theta-tan \theta)+\displaystyle\frac{b}{a}(sec \theta+tan \theta)$.

Second solution: Elegant solution: Stage 3: Two pairs of Mutually Inverse factors allowed Component expression substitution following Reduction in number of variables technique

Reduction in number of variables technique states,

In simplifying an expression, the more we are able to reduce the number of variables in the expression, easier and simpler would be the steps to the solution.

To reduce number of variables and elements in the expression drastically, we find the situation ideal for applying component expression substitution by using an intermediate dummy variable $z=\displaystyle\frac{a}{b}(sec \theta-tan \theta)$, and reduce the number variables to just 1.

From the previous result,

$2=\displaystyle\frac{a}{b}(sec \theta-tan \theta)+\displaystyle\frac{b}{a}(sec \theta+tan \theta)$,

Or, $2=z+\displaystyle\frac{1}{z}$,

where,

$z=\displaystyle\frac{a}{b}(sec \theta-tan \theta)$.

This is in accordance of the higher level Reduction in number of variables technique in an expression. The problem now is transformed to a much simpler one, and it is a case of solving a simpler problem.

Second solution: Elegant solution: Final Stage 4: Solving a simpler problem, reverse substitution and Friendly trigonometric function pairs concepts

From the previous stage,

$2=z+\displaystyle\frac{1}{z}$,

Or, $z^2-2z+1=0$.

Or, $(z-1)^2=0$,

Or, $z=\displaystyle\frac{a}{b}(sec \theta-tan \theta)=1$, by Reverse substitution,

Or, $sec \theta - tan \theta=\displaystyle\frac{b}{a}$.

This expression is perfect for applying the Friendly trigonometric function pairs concepts and with $sec \theta-tan \theta =\displaystyle\frac{1}{sec \theta + tan \theta}$ we get,

$sec \theta - tan \theta=\displaystyle\frac{b}{a}$,

Or, $\displaystyle\frac{1}{sec \theta + tan \theta} = \displaystyle\frac{b}{a}$,

Or, $sec \theta + tan \theta = \displaystyle\frac{a}{b}$.

This is familiar grounds and we just substract the value of $sec \theta -tan \theta$ from $sec \theta + tan \theta$ to get $tan \theta$,

$2tan \theta = \displaystyle\frac{a}{b} - \displaystyle\frac{b}{a}$,

Or, $tan \theta =\displaystyle\frac{1}{2ab}(a^2-b^2)$.

We always need squaring an expression when we have to derive any trigonometric function from any other, excluding the mutually inverse functions such as, $cosec \theta$ from $\sin \theta$. But we have delayed the squaring till the very last stage, and that too avoided the squaring of the functions.

Just like delayed evaluation technique, the delayed raising of power technique states,

The earlier an expression or the variables are raised in increasing power, the more cumbersome, time-consuming and complex will be the steps to the solution, and so conversely, the more raising of increasing power is delayed or avoided altogether, simpler and faster will be the steps to the solution.

Answer: d: $\displaystyle\frac{1}{2ab}(a^2-b^2)$.

In this elegant solution, we have reduced the number of variables to just one using simple algebraic manipulations. Finally, we avoided squaring of trigonometric functions completely.

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.

Key concepts and techniques used: Problem analysis -- Key pattern identification -- Principle of harmony or dicordance in expressions -- Discordant variable associations -- Friendly trigonometric function pairs concepts -- Mutually inverse expression resource -- Principle of collection of friendly terms -- Component expression substitution -- Reduction in number of variables technique -- Solving a simpler problem -- Reverse substitution technique -- Basic algebraic concepts -- Rich algebraic techniques -- Minimum order simplest solution principle -- Delayed raise in powers technique -- Efficient simplification -- Many ways technique - Real life problem solving -- Domain mapping -- Multilevel abstraction -- Degree of abstraction.

Special note

Though the detailed explanations in the elegant solution section seem to be long and abstract, the actual solution could be achieved wholly mentally. This was possible because the actions were all guided by conceptual analysis, and were inherently simple. The structures manipulated became more and more harmonious and balanced (and that can be remembered easily) after each step, converging finally to the simplest form.

If one is practiced and conversant with not only applying the powerful topic related problem solving concepts, strategies and techniques, but also analyzing a new situation to sense the right path by forming and following more general principles, no problem should pose any difficulty.

On Principle of harmony or discordance, a very general problem solving resource

The key breakthrough was achieved by applying principle of harmony or discordance which tells about harmonious relationships between various elements in an expression. Its application originated (as far as we are concerned) when solving a Trigonometric problem elegantly. Inherently though this principle belongs to the topic of Algebra where more of its applications should be found.

On afterthought though, we classify this principle as one of the most general and so, powerful principles that can be applied not only to maths, but to all kinds of problem states including real life problem solving.

In a real life problem situation, the variables and coefficients may be thought as equivalent to persons, attributes and things, whereas the associations equivalent to relations between the real life agents and elements. Obviously the binary or unary math operations may need also to be mapped suitably.

This is domain mapping by abstraction of highest order. Freedom from domain is achieved by extensive multilevel abstraction, and without freeing a problem solving principle from the topic area or domain where it is first applied, it cannot be generalized and applied in solving other domain problems.

The more is the degree of abstraction of a problem solving resource, the more is the generalization and broader is its scope of application.


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 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 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

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 elegant solutions of SSC CGL problems