**Initial problem solving strategy formulation and strategy review midway are the key steps **

Some problems look difficult, but if you have a clear strategy of dealing with this type of problems, and is ready to change track by reevaluating the strategy midway, chances are high that you will reach the efficient solution in a few quick steps.

In this session we will showcase a trigonometric expression minimization problem that looks difficult. How problem analysis, problem solving strategy formulation and review of the strategy midway to change track and use of a general more basic technique results in quick solution will be highlighted. Additionally we will apply the Many ways technique to solve the problem in two more ways.

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 minimum value of $\sin^2 \theta + \cos^2 \theta +\sec^2 \theta +\text{cosec}^2 \theta+\tan^2 \theta +\cot^2 \theta$ is,

- 3
- 5
- 1
- 7

**Solution 1 - Problem analysis and strategy formulation**

The expression involved is a long one with many terms in trigonometric functions.

We know, the most important initial objective of minimization of any trigonometric expression is,

To simplify the expression to at most a three term two function expression.

The third term may be a numeric term. This is because, such an expression has more than one method of solution with respect to minimization or maximization of the expression. The methods may be applied individually or together, forming a hybrid method.

Analyzing the given expression we decide as a strategy to simplify the expression in terms of $\tan \theta$ and $\cot \theta$. Being an inverse function pair with product 1, such an expression can easily be solved using AM GM inequality technique.

#### Solution 1 - Problem solving execution stage 1

The given expression is,

$\sin^2 \theta + \cos^2 \theta +\sec^2 \theta +\text{cosec}^2 \theta+\tan^2 \theta +\cot^2 \theta$

$=1+ (1+\tan^2 \theta) + (1+\cot^2 \theta)+\tan^2 \theta +\cot^2 \theta$

$=3+2(\tan^2 \theta +\cot^2 \theta)$.

Though we can proceed with our earlier approach and use the standard* AM GM inequality technique* to solve the problem at this point, we reviewed the problem state and decided that the more basic * algebraic approach of minimization for quadratic equations* should provide us the solution in a more elegant way.

#### Solution 1 - Intermediate stage strategy change

By the * Algebraic minima technique* a quadratic expression is converted to the following form,

$a+(b-c)^2$, where the minima $a$ will occur for $b=c$.

Otherwise, $(b-c)^2$ will always add a positive value to $a$.

This is a purely algebraic minima determination technique and more basic in nature. We can apply this technique here because the two square terms being inverse functions, their product, which will be part of the middle term of the subtractive square, will be a numeric term.

So we have the given expression transformed as,

$3 + 2(\tan^2 \theta +\cot^2\theta)$

$=7+2(\tan \theta - \cot\theta)^2$

This will have the minimum value of 7 when $\tan \theta=\cot \theta$.

**Answer:** d: 7.

**Key concepts and techniques used:** * Problem analysis* --

**--**

*Problem solving strategy formulation**--*

**Intermediate stage strategy review**

**Algebraic minima technique -- Inverse trigonometric functions -- Basic algebraic concepts -- Basic trigonometry concepts -- Maxima minima for trigonometric expressions.**

**Solution 2 - Using AM GM inequality**

In this solution path, we will proceed the same way as before to simplify the given expression as,

$\sin^2 \theta + \cos^2 \theta +\sec^2 \theta +\text{cosec}^2 \theta+\tan^2 \theta +\cot^2 \theta$

$=3 +2(\tan^2 \theta + \cot^2 \theta)$.

But at this point we decide to apply the * AM GM inequality technique* to find the minimum value of $(\tan^2 \theta + \cot^2 \theta)$.

$\text{AM}=\displaystyle\frac{\tan^2 \theta + \cot^2 \theta}{2}$.

$\text{GM}=\sqrt{\tan^2 \theta \times{\cot^2 \theta}}=1$.

By AM GM inequality concept,

$\text{AM} \geq \text{GM}$,

Or, $\displaystyle\frac{\tan^2 \theta + \cot^2 \theta}{2} \geq 1$,

Or, $\tan^2 \theta + \cot^2 \theta \geq 2$.

So the minimum value of $(\tan^2\theta +\cot^2\theta)=2$.

Thus minimum value of our given expression is,

$=3+2\times{2}=7$.

**Solution 3 - Problem analysis and solving**

In this third solution path, without any specific problem oriented strategy we decide to simplify as we can, intending it to simplify in terms of $\sin \theta$ and $\cos \theta$.

Proceeding as before we arrive at the same middle stage of simplification quickly,

$\sin^2 \theta + \cos^2 \theta +\sec^2 \theta +\text{cosec}^2 \theta+\tan^2 \theta +\cot^2 \theta$

$=3 +2(\tan^2 \theta + \cot^2 \theta)$

$=3+2\left(\displaystyle\frac{\sin^2 \theta}{\cos^2 \theta} + \displaystyle\frac{\cos^2 \theta}{\sin^2 \theta}\right)$

$=3 +2\left(\displaystyle\frac{\sin^4 \theta + \cos^4 \theta}{\sin^2 {\theta}\cos^2 \theta}\right)$

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

$=-1+\displaystyle\frac{2}{\sin^2 {\theta}\cos^2 \theta}$.

Now we will use a rich maximization concept for the denominator of the second term.

As maximum value of $\sin^n {\theta}\cos^n \theta$ is $\left(\frac{1}{2}\right)^n$, the maximum value of $\sin^2 {\theta}\cos^2 \theta$ is,

$\left(\displaystyle\frac{1}{2}\right)^2=\displaystyle\frac{1}{4}$.

Thus the minimum value of the given expression is,

$-1+2\times{4}=7$.

This is rather a long solution.

**Important**

The alternative solutions 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 we start solving the problem with a specific suitable strategy, midway through the solution, if opportunity arises, we must be ready and alert to use a new better path of solution adjusting the technique on the way.
- Any random approach will invariably take you along a confusing and longer path to the solution.
- Specifically for this type of problem, we need to adopt a strategy of simplification of the large expression into at most a three term two function expression with the two functions preferably be an inverse function pair such as $\tan$-$\cot$.

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

### 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 concepual 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 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 elega****n****t solutions of SSC CGL problems**