How to solve not so difficult SSC CGL level problem in a few light steps, Trigonometry 9

The medium level problem can be solved in just a few steps if you discover the concept based key pattern

How to solve SSC CGL level problem in quick steps trigonometry 9

The chosen problem this time is not particularly difficult, and on most occasions it will be solved conventionally. But driven by the general objective of equalizing the denominators, if the lightly hidden key pattern is discovered, everything becomes easy and straightforward with solution only a few light steps away.

In this session we will showcase a not so difficult problem. With such problems on most occasions 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.

In contrast to the conventional solution, we will present the elegant solution next. It will highlight, how driven by a general problem solving objective, a key pattern and conceptual method to use the pattern can be discovered for solving the problem in only a few easy steps.

The concepts and techniques used in the elegant solution are easy and straightforward, and anyone can think in this way, saving valuable seconds in comparison to the conventional way to the 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.

The value of $sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right) - 2tan^2 \theta$ is,

  1. 4
  2. 0
  3. 2
  4. 1

First solution: Conventional approach with deductive steps

If the two balanced terms within the brackets are combined, the numerator is formed as sum of two squares. The expression being not very complex and balanced in nature, it should not be difficult to simplifiy the sum of squares numerator. This is the conventional thinking adopted most of the times for this problem.

The given expression is,

$E=sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right) - 2tan^2 \theta$

$=sec \theta\left(\displaystyle\frac{(1+sin \theta)^2 +cos^2 \theta}{cos \theta(1+sin \theta)}\right) - 2tan^2 \theta$

$=sec \theta\left(\displaystyle\frac{1 + 2sin \theta +sin^2 \theta +cos^2 \theta}{cos \theta(1+sin \theta)}\right)-2tan^2 \theta$

$=sec \theta\left(\displaystyle\frac{2 + 2sin \theta}{cos \theta(1+sin \theta)}\right) - 2tan^2 \theta$

$=2sec \theta\left(\displaystyle\frac{(1 + sin \theta)}{cos \theta(1+sin \theta)}\right)-2tan^2 \theta$

$=2sec^2 \theta - 2tan^2 \theta$

$=2$.

Answer: Option c: 2.

It is a purely deductive solution usually followed in schools.

We will now present the elegant solution where hidden simplifying patterns are discovered and used, thus simplifying the expression significantly in just a few steps avoiding the cumbersome, time-consuming squaring and simplification process.

Second solution: Elegant Solution: Denominator equalization by denominator rationalization

Though simple in nature, in any problem involving simplification of multiple additive fractional terms, denominator equalization technique always achieves an elegant solution. In this case also, we look for an way to equalize the denominators of the two terms.

A second great result we have experienced so often,

If you look for some special result in particular, chances will be high that you will get it.

Conversely,

If you don't look for a new opportunity, you will never be able to discover it.

In this problem, when we searched for an way to equalize the denominators, the lightly hidden opportunity of rationalizing the denominator of the second term revealed itself.

Let us show how.

The given expression,

$E=sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right)-2tan^2 \theta$

$=sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta(1-sin \theta)}{1-sin^2 \theta}\right)-2tan^2 \theta$

$=sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta(1-sin \theta)}{cos^2 \theta}\right)-2tan^2 \theta$

$=sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{1-sin \theta}{cos \theta}\right)-2tan^2 \theta$

$=2sec \theta\left(\displaystyle\frac{1}{cos \theta}\right)-2tan^2 \theta$

$=2sec^2 \theta - 2tan^2 \theta$

$=2$.

Answer: Option c: 2.

Key concepts used: Deductive reasoning to attach priority to denominator equalization as a means to elegant simplification and identification and application of subsequent problem solving steps in a chained sequence -- Denominator equalization -- Key patttern identification -- Denominator rationalization -- Basic trigonometry concepts -- Basic algebra concepts -- Rich algebra techniques -- Efficient simplification -- Many ways technique.

The squaring and simplification of the denominators is completely avoided which is the hallmark of problem solver's elegant solution.

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

Important to note

Elegant and efficient solutions are based on problem analysis, deductive reasoning and use of problem solving strategies and techniques. If you attempt to solve a problem using concepts in a few steps, gradually you will develop the skillset in reaching an elegant solution for most problems easily.

It is in a way, a habit of thinking in new ways.

Conversely, if you always follow the conventional path to the solution of a problem, you will never know how to think new and how to solve problems elegantly and efficiently.

Special characteristics of efficient trigonometry problem solving

  1. To achieve elegant and efficient solutions in Trigonometry consistently, role of basic and rich algebraic concepts and techniques is critically important.
  2. While solving the problem in two 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

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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

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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

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