## 40th SSC CGL level Solution Set, topic Trigonometry 4

This is the 40th solution set for the 10 practice problem exercise for SSC CGL exam and 4th on topic Trigonometry. You may refer to the * 40th SSC CGL question set and 4th on Trigonometry* before going through this solution.

We repeat the method of taking the test. It is important to follow result bearing methods even in practice test environment.

### Method of taking the test for getting the best results from the test:

**Before start,**go throughor any short but good material to refresh your concepts if you so require.**Tutorial on Basic and rich concepts in Trigonometry and its applications****Answer the questions**in an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes.**When the time limit of 12 minutes is over,**mark up to which you have answered,**but go on to complete the set.****At the end,**refer to the answers given at the end to mark your score at 12 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time.**Identify and analyze**the problems that**you couldn't do**to learn how to solve those problems.**Identify and analyze**the problems that**you solved incorrectly**. Identify the reasons behind the errors. If it is because of**your shortcoming in topic knowledge**improve it by referring to**only that part of concept**from the best source you can get hold of. You might google it. If it is because of**your method of answering,**analyze and improve those aspects specifically.**Identify and analyze**the**problems that posed difficulties for you and delayed you**. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques.**Give a gap**before you take a 10 problem practice test again.

Important:bothandpractice testsmust be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance.mock tests

**Resources that should be useful for you**

#### Before taking the test it is recommended that you refer to

**Tutorial on Basic and rich concepts in Trigonometry and its applications.**

**Tutorial on Basic and rich concepts in Trigonometry part 2, proof of compound angle functions**

**Tutorial on Trigonometry concepts part 3, finding maxima or minima of Trigonometric expressions**

**You may also refer to the related resources:**

* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****generally for any hard MCQ test.**

If you like,you mayto get latestsubscribecontent on competitive examspublished in your mail as soon as we publish it.

### 40th solution set- 10 problems for SSC CGL exam: 4th on Trigonometry - test time 12 mins

**Problem 1.**

The $cosec 39^0=p$, the value of, $\displaystyle\frac{1}{cosec^2 51^0} + sin^2 39^0 + tan^2 51^0 - \displaystyle\frac{1}{sin^2 51^0 sec^2 39^0}$ is,

- $p^2 - 1$
- $\sqrt{p^2 - 1}$
- $1-p^2$
- $\sqrt{1-p^2}$

**Solution - Problem analysis and execution**

From * Complementary Trigonometric functions concepts* we know,

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

$cos\left(\displaystyle\frac{\pi}{2} - \theta\right) = sin\theta$, and so,

$tan\left(\displaystyle\frac{\pi}{2} - \theta\right) = cot\theta$, where $\theta$ is acute.

With this knowledge and analyzing the terms of the target expression we decide to transform the requisite functions giving the target expression,

$E=sin^2 51^0 + sin^2 39^0 + tan^2 51^0 - \displaystyle\frac{cos^2 39^0}{sin^2 51^0}$

$\hspace{5mm}=cos^2 39^0 + sin^2 39^0 + cot^2 39^0 - \displaystyle\frac{cos^2 39^0}{cos^2 39^0}$

$\hspace{5mm}=1 + (cosec^2 39^0 - 1) - 1$

$\hspace{5mm}=p^2 - 1$

**Answer:** a: $p^2 - 1$.

Before converting the terms to their complementary form, we have analyzed the potential of simplification and then only converted the functions into complementary functions so that the basic trigonometric concepts such as $sin^2 \theta + cos^2 \theta=1$ and $cot^2 \theta = cosec^2 \theta - 1$ can be used for faster simplification.

**Key concepts and techniques used:** * Identifying the useful pairing patterns* in the series of terms --

*-- use of*

**Key pattern identification***for $sin\theta$, $cos\theta$ and $tan \theta$ -- Basic and Rich Trigonometric Concepts.*

**Complementary Trigonometric functions concepts****Note:** In this case the analysis and simplifying actions went hand in hand wholly in mind producing the final result in no time. Only the initial identification of the patterns of result bearing terms was needed.

**Problem 2.**

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

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

**Solution - Problem analysis and execution**

We identify the value of $sec \theta$ as similar to the form of sum of inverses we have encountered so often in Algebraic problem simplification where * Principle of interaction of inverses* led us to the desired solution always. We also recognize that for using the method we need to make minor

**transformation of the input expression.**#### Solution - Simplifying actions

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

Or, $sec \theta=\displaystyle\frac{1}{2}\left(2x + \displaystyle\frac{1}{2x}\right)$.

As we know, $sec^2 \theta -1 = tan^2 \theta$, squaring and subtracting 1 from both sides of the above equation we get,

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

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

$\hspace{19mm}=\displaystyle\frac{1}{4}\left(2x - \displaystyle\frac{1}{2x}\right)^2$,

Or, $tan \theta = \displaystyle\frac{1}{2}\left(2x - \displaystyle\frac{1}{2x}\right)$.

Adding $sec \theta$ and $tan \theta$ expressions finally we get,

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

**Answer:** b: $2x$.

**Key concepts and techniques used:** Identification and use of * sum of inverses pattern* --

*--*

**principle of interaction of inverses***of sum of inverses -- basic trigonometric concepts.*

**input transformation****Problem 3.**

If $tan \theta=1$, then the value of $\displaystyle\frac{8sin \theta + 5cos \theta}{sin^3 \theta -2cos^3 \theta + 7cos \theta}$ is,

- $2\displaystyle\frac{1}{2}$
- $2$
- $3$
- $\displaystyle\frac{4}{5}$

**Solution - Problem analysis**

First observation is, the target expression contains only $sin \theta$ and $cos \theta$, whereas the value of $tan \theta$ is given in simplest terms of 1. This urges us to transform the given exptression as,

$tan \theta = 1$,

Or, $sin \theta = cos \theta$.

Because of the above reasoning, this result should be the key to simplify the target expression quickly and elegantly.

**Solution - Simplifying actions**

So the target expression can be expressed now as,

$E=\displaystyle\frac{8sin \theta + 5sin \theta}{sin^3 \theta - 2sin^3 \theta + 7sin \theta}$

$\hspace{5mm}=\displaystyle\frac{13}{-sin^2 \theta + 7}$.

At this point we realize that we also have to evaluate $sin^2 \theta$ from the input value $tan \theta =1$. This is the * second way to use the same input expression.* Thus we have,

$tan \theta = 1$,

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

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

Or, $sin^2 \theta = \displaystyle\frac{1}{2}$.

Substituting,

$E=\displaystyle\frac{13}{-\frac{1}{2} + 7}$

$\hspace{5mm}=\displaystyle\frac{13}{\displaystyle\frac{13}{2}}$

$\hspace{5mm}=2$

**Answer:** b: 2.

**Key concepts and techniques used:** * End state analysis* --

*-- simplification --*

**Target based input transformation***(we have used the input expression in two ways).*

**trigonometric basic function transformation principle -- Multiple input use technique**Multiple input use is a powerful technique.

#### Trigonometric basic function tranformation principle:

If value of one of the trigonometric functions, $sin \theta$, $cos \theta$, $tan \theta$, $cosec \theta$, $sec \theta$ or $cot \theta$ is given, any of the other functions can be derived using the following basic trigonometric relations,

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

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

$cosec^2 \theta=1 + cot^2 \theta$ and the inverse functions of,

$cosec \theta = \displaystyle\frac{1}{sin \theta}$,

$sec \theta = \displaystyle\frac{1}{cos \theta}$ and

$cot \theta = \displaystyle\frac{1}{tan \theta}$.

**Problem 4.**

If $7sin \theta = 24cos \theta$, where $0 \lt \theta \lt \displaystyle\frac{\pi}{2}$, then the value of $14tan \theta - 75cos \theta - 7sec \theta$ is,

- 1
- 3
- 2
- 4

**Solution - Problem analysis**

By * End state analysis,* comparing the target expression with the input expression we find that the value of $tan \theta$ in the target expression can easily be derived from the input using direct transfomation of the input,

$7sin \theta = 24cos \theta$,

Or, $tan \theta = \displaystyle\frac{24}{7}$.

This leaves basically $cos \theta$ in the target expression. So we need a second input tranformation applying **Multiple input use technique,**

$7sin \theta = 24cos \theta$,

Or, $tan \theta = \displaystyle\frac{24}{7}$,

Or, $tan^2 \theta + 1 = \displaystyle\frac{24^2}{7^2} + 1$,

Or, $sec^2 \theta = \displaystyle\frac{576}{49} + 1=\displaystyle\frac{625}{49}$,

Or, $sec \theta = \displaystyle\frac{25}{7}$, as $0 \lt \theta \lt \displaystyle\frac{\pi}{2}$, $sec \theta$ is positive.

So,

$cos \theta = \displaystyle\frac{7}{25}$.

#### Solution - Simplifying actions

Using these values in the target expression we have,

$E=14tan \theta - 75cos \theta - 7sec \theta$,

$\hspace{5mm}=14\times{\displaystyle\frac{24}{7}} - 75\times{\displaystyle\frac{7}{25}} -7\times{\displaystyle\frac{25}{7}}$

$\hspace{5mm}=48 - 21 - 25$

$\hspace{5mm}=2$.

**Answer:** c: 2.

**Key concepts and techniques used:** *End state analysis* -- input transformation technique -- * Multiple input use technique* -- basic and rich trigonometric concepts --

**Trigonometric basic function transformation principle.****Problem 5.**

The minimum value of $sin^2 \theta + cos^2 \theta + sec^2 \theta + cosec^2 \theta + tan^2 \theta + cot^2 \theta$ is equal to,

- 1
- 7
- 3
- 5

**Solution - Problem analysis**

On quick analysis we can visualize the first two terms being transformed to a numeral and the last four terms resulting in a part expression that is a sum of square of inverses. Let us see how.

Transforming the target expression thus we have,

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

$\hspace{5mm}=3 + 2(tan^2 \theta + cot^2 \theta)$.

This expression potentially conforms to the well known algebraic technique for finding maxima and minima in quadratic expressions.

#### Maxima minima technique

**Problem:**

Find the minimum value (or minima) of the expression, $ax^2 + bx + c$, where variable $x$ has real values.

**Solution: **

The standard technique to solve this problem is to transform the expression so that one part of it becomes a square of sum of two terms where all the variable terms are absorbed in the square expression leaving only one numeric term outside the square expression, like,

$ax^2 + bx + c = p + (qx - r)^2$.

From this transformed expression we can then mathematically conclude that, for all real values of variable $x$, the square term will always be positive and hence will increase the value of the expression from the minimum value of $p$ except when the square term is zero.

#### Problem solving execution

With this knowledge the target expression is transformed to the desired form as,

$E=3 + 2(tan^2\theta -2 + cot^2\theta) + 4$

$\hspace{5mm}=7+(tan \theta - cot\theta)^2$.

The square term of this expression will always be positive except when $tan\theta=cot\theta$.

Thus the minimum value of the given expression is 7.

**Answer:** b: 7.

**Key concepts and techniques used:** Basic trigonometric concepts* -- *simplification

**-- transformation of the square variable terms into a square of a two term expression -- algebraic maxima minima technique - rich algebraic concepts.**

**Problem 6.**

In a right $\triangle ABC$ with right angle at $\angle ABC$, if $AB=2\sqrt{6}$ and $AC - BC = 2$ then, $sec A + tan A$ is,

- $\displaystyle\frac{\sqrt{6}}{2}$
- $2\sqrt{6}$
- $\sqrt{6}$
- $\displaystyle\frac{1}{\sqrt{6}}$

**Solution - **Problem analysis

The following figure describes the problem,

As the length of a side is given and the target expression is in usual form of sum of basic trigonometric functions that primarily are ratios of side lengths of a right triangle, we decide to go for finding the length of sides using Pythagoras theorem rather than trigonometric simplification of target expression.

#### Solution - Simplifying actions

In $\triangle ABC$,

$AB = 2\sqrt{6}$ and,

$AC - BC=2$.

Multiplying both sides by $(AC+BC)$ we get,

$AC^2 - BC^2 = 2(AC+BC)$,

Or, $AB^2 = 2(AC+BC)$,

Or, $AC + BC = \frac{1}{2}(2\sqrt{6})^2 = 12$.

Adding this with the equation of $AC - BC =2$ we get,

$2AC = 14$,

Or, $AC=7$.

And subracting,

$2BC = 10$,

Or, $BC=5$.

So,

$sec A = \displaystyle\frac{AC}{AB} = \displaystyle\frac{7}{2\sqrt{6}}$, and

$tan A = \displaystyle\frac{BC}{AB} = \displaystyle\frac{5}{2\sqrt{6}}$, giving,

$sec A + tan A = \displaystyle\frac{12}{2\sqrt{6}}=\sqrt{6}$.

**Answer:** c: $\sqrt{6}$.

**Key concepts and techniques used:** * Deductive reasoning* --

*-- adopting the*

**End state analysis approach***-- basic algebraic concept of $(a+b)(a-b)=a^2-b^2$ --*

**strategy of finding the length of sides***-- basic algebraic concept of solving two linear equations in two variables -- basic trigonometric functions as ratio of sides of a right triangle -- basic trigonometric concepts.*

**Pythagoras theorem****Problem 7.**

If $tan 2\theta . tan 4\theta = 1$, then the value of $tan 3\theta$ is,

- $\sqrt {3}$
- $0$
- $\displaystyle\frac{1}{\sqrt{3}}$
- $1$

**Solution - problem analysis**

By comparing the target expression with the input expression we don't find any other solution path than to find the actual value of the angle $\angle \theta$ itself.

**Solution - simplifying actions**

$tan 2\theta . tan 4\theta = 1$,

Or, $tan 2\theta = \displaystyle\frac{1}{tan 4\theta}=cot 4\theta$.

From * Complementary Trigonometric functions concepts* we know,

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

$cos\left(\displaystyle\frac{\pi}{2} - \theta\right) = sin\theta$, and so,

$tan\left(\displaystyle\frac{\pi}{2} - \theta\right) = cot\theta$, where $\theta$ is acute.

So,

$tan 2\theta = cot 4\theta=tan\left(\displaystyle\frac{\pi}{2} - 4\theta\right)$,

Or, $2\theta = \left(\displaystyle\frac{\pi}{2} - 4\theta\right)$,

Or, $6\theta = \displaystyle\frac{\pi}{2}$,

Or, $3\theta = \displaystyle\frac{\pi}{4}=45^0$, and

$tan 3\theta=tan 45^0=1$.

**Answer:** d: 1.

**Key concepts used:** Deductive reasoning -- adopting the strategy of finding the value of the angle $\theta$ itself -- use of * Complementary Trigonometric functions concepts *for $tan\theta$ -- final simplification.

**Problem 8.**

If $sin \displaystyle\frac{\pi x}{2}=x^2 -2x +2$, then the value of $x$ is,

- $0$
- $-1$
- $1$
- none of these

**Solution - problem analysis**

Getting no direct path to the solution the obvious decision has been to examine the RHS expression closely.

**Solution - Problem solving execution**

$sin \displaystyle\frac{\pi x}{2}= x^2 - 2x + 2 = (x-1)^2 + 1$.

As for any real value of $x$ the square of sum term will be positive it will make the value of the RHS larger than 1 which is not possible for any value of a $sin$ function. So the square term must be 0 and so, $x=1$.

**Answer:** c: 1.

**Key concepts and techniques used:** Deductive reasoning -- * analysis of RHS to transform it to a square of sum term plus 1 form* -- basic trigonometric concepts -- basic algebraic concepts.

**Problem 9.**

If $2sin \theta + cos \theta = \displaystyle\frac{7}{3}$, then the value of $(tan^2 \theta - sec^2 \theta)$ is,

- $0$
- $\displaystyle\frac{7}{3}$
- $\displaystyle\frac{3}{7}$
- $-1$

**Solution** - Problem analysis and execution

As a rule we examine the target expression and this time it handed us the solutuon in one step without going into any complexity of using the awkward input expression,

$(tan^2 \theta - sec^2 \theta)=tan^2\theta - (1 + tan^2 \theta)=-1$

**Answer:** d: $-1$.

**Key concepts and techniques used:** Target expression analysis -- basic trigonometric concepts.

**Problem 10.**

If $(rcos \theta - \sqrt{3})^2 + (rsin \theta - 1)^2 = 0$, then the value of $\displaystyle\frac{rtan \theta + sec \theta}{rsec \theta + tan \theta}$ is,

- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{\sqrt{3}}{4}$
- $\displaystyle\frac{\sqrt{5}}{4}$
- $\displaystyle\frac{5}{4}$

#### Solution - Problem analysis

While analysing the problem the very first thing that strikes us is the presence of zero sum of square terms. In such situations we know that individually the square terms must be zero. So we have the breakthrough at the very beginning.

From the given expression,

$(rcos \theta - \sqrt{3})=0$,

Or, $rcos \theta = \sqrt{3}$, and

$(rsin \theta -1)=0$,

Or, $rsin \theta = 1$.

Because of our awareness of the values of $sin \theta$ and $cos \theta$ for various values of $\theta$, it is recognized immediately that the often used values of,

$sin30^0 = \displaystyle\frac{1}{2}$, and

$cos30^0 = \displaystyle\frac{\sqrt{3}}{2}$

will satisfy the two equations nicely, giving,

$r=2$, and

$\theta=30^0$.

#### Solution - Derivation and Simplifying actions

To derive the above results we square the two expressions and sum up giving,

$r^2(sin^2 \theta + cos^2 \theta) = r^2 = 4$,

Or, $r=2$.

Substituting this value of $r$ in any of the two equations we get the value of $\theta$ as $30^0$.

Using these values of $r=2$ and $\theta=30^0$ in the target expression we have the simplified target expression as,

$E=\displaystyle\frac{2tan 30^0 + sec 30^0}{2sec 30^0 + tan 30^0}$

$\hspace{5mm}=\displaystyle\frac{\displaystyle\frac{2}{\sqrt{3}} + \displaystyle\frac{2}{\sqrt{3}}}{\displaystyle\frac{4}{\sqrt{3}} + \displaystyle\frac{1}{\sqrt{3}}}$

$\hspace{5mm}=\displaystyle\frac{\displaystyle\frac{4}{\sqrt{3}}}{\displaystyle\frac{5}{\sqrt{3}}}$

$\hspace{5mm}=\displaystyle\frac{4}{5}$.

**Answer:** a: $\displaystyle\frac{4}{5}$.

**Key concepts and techniques used:** *Rich algebraic principle of zero sum of square terms -- basic algebraic techniques to find values of $r$ and $\theta$ -- substitution and simplification.*

**Note:** You will observe that in many of the Trigonometric problems rich algebraic concepts and techniques are to be used. In fact that is the norm. Algebraic concepts are frequently used for elegant solutions of Trigonometric problems.

### 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**

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**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**

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**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.

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