**Rich trigonometry concept of Friendly Function Pair enables elegant solution **

In * Basic and Rich Trigonometry concepts and applications*, we have explained how rich problem solving trigonometry concepts are derived from basic concepts for solving problems faster. In this session we will highlight

- how the rich trigonometry concept of
is derived from basic trigonometry concepts, and**Friendly Trigonometric Function Pair** - how the rich concept thus derived can be used for solving different and difficult SSC CGL level problems in only a few steps elegantly.

We will use two specially selected problems to show the power of the rich trigonometry concept, that essentially is a powerful mathematical problem solving technique derived from very basic concepts in Trigonometry.

**Note:** we have used this concept in an * earlier session* but didn't explore it fully leaving it without naming and without showing its applicatibility in solving varieties of problems.

We will now define the rich concept in more concrete terms first.

**The rich concept of Friendly Trigonometric Function Pair**

As one of the important rich concepts, we recognize three pairs of trigonometric functions as * friendly trigonometric function pairs*. These are,

$sin \theta$ and $cos \theta$,

$sec \theta$ and $tan \theta$, and

$cosec \theta$ and $cot \theta$.

The most basic relationship between two trigonometric functions involves the first friendly function pair of $sin \theta$ and $cos \theta$,

$sin^2 \theta + cos^2 \theta=1$.

This is one of the most frequently used trigonometric problem solving resource that we know. We have classified this pair as a friendly function pair because of the intimate relationship between them and the effectiveness of the relation to simplify trigonometry problems time and again.

We will only list out the various forms in which the relation is used and move on to the other two pairs of functions,

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

$sin^2 \theta = 1-cos^2 \theta$,

$cos^2 \theta = 1-sin^2 \theta$.

The structure of the basic relationship and its use for this most used function pair are a little different from that of the the other two pairs. We will leave the relationship in these forms as essentially basic trigonometry concepts. Nevertheless because of the intimate ties between the two functions we classify them also under the friendly function pair without adding any rich functionalty to this pair of functions.

Instead we will add rich functionality to the other two function pairs, namely, $sec \theta$ and $tan \theta$; and $cosec \theta$ and $cot \theta$ which together act as the problem solving content of this powerful concept of friendly trigonometric function pair.

#### The highly effective problem solving relationship between the elements of a friendly function pair

The basic relationship between the first function pair of $sec \theta$ and $tan \theta$ is,

$sec^2 \theta = 1+tan^2 \theta$.

This is the form in which this function pair is mostly used. In this form the relationship is a basic trigonometry concept. We will transform the relationship to a different form.

$sec^2 \theta = 1+tan^2 \theta$,

Or, $sec^2 \theta -tan^2 \theta=1$,

Or, $(sec \theta +tan \theta)(sec \theta -tan \theta)=1$,

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

Alternately,

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

The inverse relationship between the two additive and subtractive complementary expressions of the two functions lends the power to solve complex problems elegantly.

In the same way we recognize the inverse relationship between $cosec \theta$ and $cot \theta$,

$cosec^2 \theta = 1+cot^2 \theta$,

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

Or, $(cosec \theta + cot \theta)(cosec \theta - cot \theta)=1$,

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

Alternately,

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

We will now show how these enriched relationships can be used for solving complex trigonometry problems in only a few steps elegantly.

**Chosen Problem 1.**

If $a=\sec \theta+\tan \theta$, then $\displaystyle\frac{a^2-1}{a^2+1}$ is,

- $\sec \theta$
- $\cos \theta$
- $\tan \theta$
- $\sin \theta$

#### Solution to chosen problem 1 - Problem analysis

Not only do we recognize the presence of a * friendly trigonometric function pair* in the given input function, we also identify in the target expression, the potential to apply the powerful rich algebraic technique of

**Componendo and Dividendo.**#### The rich algebraic technique of componendo dividendo

We apply this technique on an expression of the form,

$\displaystyle\frac{x+y}{x-y}=\frac{p}{q}$, where $x$ and $y$ are the unknown variables and $p$ and $q$ are usually constants.

As a first step we add 1 to both sides of the equation giving,

$\displaystyle\frac{2x}{x-y}=\frac{p+q}{q}$.

In the second step we subtract 1 from both sides of the original equation giving,

$\displaystyle\frac{2y}{x-y}=\frac{p-q}{q}$.

In the third step we take the ratio of the two results to get,

$\displaystyle\frac{x}{y}=\frac{p+q}{p-q}$.

The LHS being a simple ratio of the two variables and the RHS involving usually numbers, the result simplifies the problem significantly.

Returning to our problem we observe, though the target expression form is perfectly amenable for applying the technique of componendo and dividendo, the final hurdle to the elegant solution turns out to be the fact that the input expression is in terms of variable $a$ while the target expression involves variable $a^2$.

To overcome this barrier we would now transform the variable $a$ to $a^2$ by applying the rich concept of friendly trigonometric function pair,

$a=sec\theta+tan\theta=\displaystyle\frac{1}{sec\theta -tan\theta}$.

Multiplying the two equations,

$a^2=\displaystyle\frac{sec\theta + tan\theta}{sec\theta -tan\theta}$.

Now it is only a simple step to the solution.

#### Solution to chosen problem 1 - Problem solving final stage

Applying the componendo dividendo technique on the equation (subtracting 1 from both sides, again adding 1 to both sides of the original equation and taking the ratio of the two),

$\displaystyle\frac{a^2-1}{a^2+1}=\frac{tan\theta}{sec\theta}=sin\theta$.

We have reached the solution in only a few steps by applying rich trigonometry * concept of friendly trigonometric function pair* as well as

**rich algebraic technique of componendo dividendo.**This is an example of achieving what we call * elegant solution* through the process of

**efficient problem solving.****Answer:** d: $sin \theta$.

**Key concepts and techniques used:** Basic trigonometry concepts* -- rich trigonomtery concepts *--

**concept of friendly trigonometric function pair****-- target driven approach in forming first $a^2$ and then the desired fraction -- rich algebra techniques -- componendo dividendo technique.**

To solve the second chosen problem we will use the rich friendly inverse relationsip between the second pair of functions $cosec \theta$ and $cot \theta$. But this time we will use the relationship in a different way to simplify the target expression quickly and elegantly.

This shows the **basic characteristic of a rich problem solving concept,**

is just a potentially powerful problem solving conceptA rich problem solving concepttothat can be used in various wayssolve many different problemshaving a common characteristic of applicability of the specific rich concept.

Let us clarify by going through the process of solving the second problem.

**Chosen Problem 2.**

The value of $\displaystyle\frac{cot \theta + cosec \theta - 1}{cot \theta -cosec \theta +1}$ is,

- $cosec \theta - cot \theta$
- $cosec \theta + cot \theta$
- $sec \theta + cot \theta$
- $cosec \theta + tan \theta$

**Solution to chosen problem 2 - **Problem analysis

Examining this problem, here also we detect the presence of a * friendly function pair*, $cosec \theta + cot \theta$. Only, we have to use this key pattern intelligently.

On further examination and comparison of the denominator and the numerator of the target expression the path to the solution becomes clear.

#### Solution to chosen problem 2 - Problem solving execution

Applying the inverse relationship of,

$cosec \theta + cot \theta =\displaystyle\frac{1}{cosec \theta - cot \theta}$, we transform the target expression to,

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

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

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

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

$=cosec \theta + cot \theta$.

Again detecting the very useful pattern of friendly trigonometric function pair and using the pattern judiciously we reach the solution elegantly in a few simple steps.

**Answer:** b: $cosec \theta + cot \theta$.

**Key concepts and techniques used:** * Key patten identification *--

*--*

**concept of friendly trigonometric function pair****-- basic algebraic concepts -- efficient simplification.**

*rich trigonometry concepts***Important:** To be able to apply a rich problem solving concept, we need to

- first, be fully aware of the concept,
- second, be aware of the basic patterns involved in the concept,
- third, detect the useful patterns in the problem, and finally,
- fourth, decide how to exploit the rich concept utilizing the power of the problem solving pattern and execute the steps.

We need not only to go through the concept and its applicability, ** we must also solve a number of problems independently** using the concept to be able to apply it when it is needed.

### 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 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 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**