## A few simple steps to the solution compared to conventional complex procedure

At high school level many times we find math problems are solved following 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 scneario, 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**.

We will take up a few 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 problems

*as well as*

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

### Problem example 1

If $\sec\theta = x + \displaystyle\frac{1}{4x}$, where $(0^0 \lt \theta \lt 90^0)$, then $\sec\theta + \tan\theta$ is,

- $\displaystyle\frac{x}{2}$
- $\displaystyle\frac{1}{2x}$
- $x$
- $2x$

**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.

We know that the basic relation between $\sec\theta$ and $\tan\theta$

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

will lead us towards the solution whenever in a problem $\sec\theta$ and $\tan\theta$ appear together.

In our problem, though the two terms appear together, these are in unit power form, and so the given expression must be squared, resulting expression simplified using the basic relationship as mentioned above, and then finally a square root is to be taken to arrive at the desired result.

This is what we call deductive reasoning based on problem analysis and using the subject concept and form of the problem.

**Outcome of this analysis is finding a clear pathway to the solution.**

Think over. Do you find this thread of reasoning sensible? Can you find any flaw in it?

So, we decide that **we should square up the given equation first.**

**Second clue - use of principle of inverses**

We find **a special property in the given expression** - it has an $x$ and also an inverse of $x$. If we square up this expression the middle term won't have any $x$ in it. This property in general helps to reach the solution quickly in so many problems that we have named it as the **powerful problem solving principle of inverses and repeatedly used it with great benefits. **

You may refer to a detailed treatment of its use **here.**

So in the first stage action, we will **first square up the given equation** and **use the principle of inverses** to simplify further.

Let's see how.

**First stage action:**

We have the given expression,

$\sec\theta = x + \displaystyle\frac{1}{4x}$.

Squaring both sides,

$\sec^2\theta = x^2 + \displaystyle\frac{1}{16x^2} + \frac{1}{2}$.

By the *grace of principle of inverses,* the middle term on the RHS has turned to a simple fraction without any trace of $x$.

Continuing further,

$\tan^2\theta + 1 = x^2 + \displaystyle\frac{1}{16x^2} + \displaystyle\frac{1}{2}$

Now we will use another great principle, the * principle of collection of friendly terms.* In the given expression we spot the possiblity of significant gains if we transfer the 1 from LHS to RHS so that the middle term now changes its sign and forms the expression of another square.

$\tan^2\theta = x^2 + \displaystyle\frac{1}{16x^2} - \displaystyle\frac{1}{2} = \left(x - \frac{1}{4x}\right)^2$,

Or, $\tan\theta = x - \displaystyle\frac{1}{4x}$, as $\tan\theta$ can't be negative as per the given condition.

Summing it up now with $\sec\theta$ from given expression,

$\sec\theta + tan\theta = 2x$.

**Answer:** d: $2x$.

#### Conventional solution

We have the given expression,

$\sec\theta = x + \displaystyle\frac{1}{4x}$.

Or squaring up the two sides of the equation we have,

$\sec^2\theta = \displaystyle\frac{(4x^2 + 1)^2}{(4x)^2}$,

Or, $\sec^2\theta - 1 = \displaystyle\frac{(4x^2 + 1)^2 - (4x)^2}{(4x)^2}$,

Or, $\tan^2\theta = \displaystyle\frac{16x^4 + 8x^2 + 1 - 16x^2}{(4x)^2}$,

$=\displaystyle\frac{16x^4 - 8x^2 + 1}{(4x)^2}$

$=\displaystyle\frac{(4x^2 -1)^2}{(4x)^2}$.

So, $\tan\theta = \displaystyle\frac{(4x^2 -1)}{(4x)} = x - \frac{1}{4x}$

and finally,

$\sec\theta + \tan\theta = 2x$.

* Compare the two solutions yourself *regarding

**ease**,

**complexity**,

**chances of error**and

**time taken**to reach the solution.

### Problem example 2

If $tan\theta = \displaystyle\frac{1}{\sqrt{11}}$, and $0 \lt {\theta} \lt \displaystyle\frac{{\pi}}{2}$, then the value of, $\displaystyle\frac{cosec^2\theta - sec^2\theta}{{cosec^2\theta} + sec^2\theta}$ is,

- $\displaystyle\frac{3}{4}$
- $\displaystyle\frac{6}{7}$
- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{5}{6}$

### Efficient solution in a few steps

#### Problem analysis

By looking at the problem, we recognize the target expression to be absolutely ready to be subjected to the well known * algebraic technique of componendo dividendo.* Though the name is a bit awkward the concept is rather easy.

We will apply the concept but not the formula here. We have a strong apathy to use formulae without using our brains.

The target expression,

$E = \displaystyle\frac{cosec^2\theta - sec^2\theta}{{cosec^2\theta} + sec^2\theta}$

We add 1 to both sides and simplify.

$E + 1 = \displaystyle\frac{cosec^2\theta - sec^2\theta}{{cosec^2\theta} + sec^2\theta} + 1$

$= \displaystyle\frac{2cosec^2\theta}{{cosec^2\theta} + sec^2\theta}$.

Second time we subtract 1 from both sides of the original equation,

$E - 1 = \displaystyle\frac{cosec^2\theta - sec^2\theta}{{cosec^2\theta} + sec^2\theta} - 1$

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

Dividing the earlier result of $E + 1$ by this result,

$\displaystyle\frac{E + 1}{E - 1} = \frac{cosec^2\theta}{-sec^2\theta} $

$= -cot^2\theta = -11$.

Adding and subtracting 1 to both sides we have,

$\displaystyle\frac{2E}{E - 1} = -10$, and

$\displaystyle\frac{2}{E - 1} = -12$.

Taking the ratio,

$E = \displaystyle\frac{10}{12} = \displaystyle\frac{5}{6}$

**Answer:** d: $\displaystyle\frac{5}{6}$.

### Cumbersome solution

In the most cumbersome solution, you can expand both the terms $cosec^2\theta = 1 + cot^2\theta$ and $sec^2\theta = 1 + tan^2\theta$ and substitute in the already complex target expression to get the target only in terms of $tan^2\theta$ and $cot^2\theta$, value of both of which are known.

We can say for this approach that, you don't have to think at all, you just have to go on deducing mechanically.

Judge and choose yourself.

**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**

### 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**