## A few simple steps to the solution

At high school level often we find that math problems are solved in a long series of steps. This is what we call * conventional approach to solving problems*.

This approach not only involves **a large number of steps**, in most cases, the steps themselves introduce a higher level of **complexity**, and **increases chances of error.*** More importantly, the conventional inefficient problem solving approach curbs the out-of-the-box thinking skills of the students.*

While dealing with complex Trigonometry problems similar to school level in competitive exam scenario, the student is now forced to solve such a problem in a minute, and not in many minutes. The pressure to find the solution along the shortest path gains immense importance for successful performance in such tests as **SSC CGL.**

Though at school level, all steps to the solution are to be written down, that does not take up most of the time, **the bulk of the time is actually consumed in inefficient problem solving, finding the path and steps to the solution - in thinking through the barriers to the solution**.

We will take up two apparently difficult Trigonometry problems from * SSC CGL test level* that actually belong to school level, and appear in MCQ form in the competitive test scenario.

The * thinking process that we will highlight *here through solution of the problem

*as well as*

**can help SSC CGL aspirants***high school students*to solve problems efficiently like a problem solver, using deductive reasoning, powerful strategies, techniques and basic subject concepts, rather than being constrained by the costly routine approach.

### Problem example 1

If $xcos\theta - ysin\theta = \sqrt{x^2 + y^2}$, and $\displaystyle\frac{cos^2\theta}{a^2} + \displaystyle\frac{sin^2\theta}{b^2}=\frac{1}{x^2 + y^2}$ then the correct relation among the following is,

- $\displaystyle\frac{x^2}{b^2} - \displaystyle\frac{y^2}{a^2} = 1$
- $\displaystyle\frac{x^2}{b^2} + \displaystyle\frac{y^2}{a^2} = 1$
- $\displaystyle\frac{x^2}{a^2} - \displaystyle\frac{y^2}{b^2} = 1$
- $\displaystyle\frac{x^2}{a^2} + \displaystyle\frac{y^2}{b^2} =1$

**First try to solve this problem yourself and then only go ahead. You might be able to reach the elegant solution to this problem yourself.**

### Efficient solution in a few steps

#### Deductive reasoning:

**First stage analysis:** one must analyze the problem first.

Our first observation is, all terms in the target expression and the second given expression are in squares. This urges us straightaway to square up the first given expression and simplify as much as possible.

This fragment of statement "as much as possible" is important that we will see.

$xcos\theta - ysin\theta = \sqrt{x^2 + y^2}$,

Or, $x^2cos^2\theta - 2xysin\theta{cos\theta} + y^2sin^2\theta = x^2 + y^2$.

This gives us the opportunity to apply the * principle of collection of friendly terms* by collecting the coefficients of $x^2$ and $y^2$ from both sides together. We would be more interested to do this because we can see that by this action, we will simplify the factors $x^2(1 - cos^2\theta)$ and $y^2(1 - sin^2\theta)$.

Rearranging the terms, taking all of them on one side of the equation we get,

$x^2(1 - cos^2\theta) + 2xysin\theta{cos\theta} + y^2(1 - sin^2\theta) = 0$,

Or, $(xsin\theta + ycos\theta)^2 = 0$

Or, $xsin\theta + ycos\theta = 0$,

Or, $\displaystyle\frac{x}{y} = -cot\theta$

Or, $\displaystyle\frac{x^2}{y^2} = cot^2\theta$.

This is what we wanted, because from our intial analysis, it was clear that we need to eliminate $sin\theta$ and $cos\theta$ and to do that we must have had a relationship of a trigonometric function in terms of $x^2$ and $y^2$.

We are happy with any trigonometric function because from basic concepts we know,

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

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

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

If we get the value of one of the basic identities we can derive any other.

To eliminate we need to get the value of $sin^2\theta$ and $cos^2\theta$ in terms of $x^2$ and $y^2$. Let's get those relations now,

$\displaystyle\frac{x^2}{y^2} = cot^2\theta = cosec^2\theta - 1$,

Or, $cosec^2\theta = \displaystyle\frac{x^2 + y^2}{y^2}$,

Or, $sin^2\theta = \displaystyle\frac{y^2}{x^2 + y^2}$, and so,

$cos^2\theta = \displaystyle\frac{x^2}{x^2 + y^2}$.

Substituting these two in the target expression we find the factor, $\displaystyle\frac{1}{x^2 + y^2}$ common to both the terms in LHS and also on the RHS and thus it cancels out, leaving just,

$\displaystyle\frac{x^2}{a^2} + \displaystyle\frac{y^2}{b^2} =1$.

**Answer:** d: $\displaystyle\frac{x^2}{a^2} + \displaystyle\frac{y^2}{b^2} =1$.

### Problem example 2

If $x = \displaystyle\frac{cos\theta}{1 - sin\theta}$, then the expression, $\displaystyle\frac{cos\theta}{1 + sin\theta}$ is,

- $\displaystyle\frac{1}{x + 1}$
- $\displaystyle\frac{1}{1 - x}$
- $\displaystyle\frac{1}{x}$
- $x - 1$

### Efficient solution in a few steps

#### Problem analysis

We find $1+sin\theta$ in the target expression whereas its complementary expression, $1 - sin\theta$ is already there in the given expression.

Why complementary? Because product of the two will be $1 - sin^2\theta = cos^2\theta$, always simplifying expressions in Trigonometry.

So we multiply both numerator and denominator of given expression by $1 + sin\theta$,

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

$\hspace{5mm}=\displaystyle\frac{cos\theta(1 + sin\theta)}{cos^2\theta}$,

$\hspace{5mm}=\displaystyle\frac{1 + sin\theta}{cos\theta}$,

So just inverting we get,

$\displaystyle\frac{cos\theta}{1 + sin\theta} = \frac{1}{x}$.

**Answer:** c: $\displaystyle\frac{1}{x}$.

Solutions to both the problems have come quickly and so we won't analyze further.

Generally we find this class of problem is solved in longer steps. It would be so because, unless you see through the barriers of the problem very quickly applying the concepts and techniques that are relevant to the problem to take you to the solution actually in a few simple steps, more often than not you will be wasting valuable time in searching for the solution.

We will end with our standard advice,

**Always think:** is there any other shorter better way to the solution? And use your brains more than your factual memory and mass of mechanical routine procedures.

### 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**