## 56th SSC CGL level Solution Set, topic Trigonometry 5

This is the 56th solution set for the 10 practice problem exercise for SSC CGL exam and 5th on topic Trigonometry.

We repeat the method of taking the test. It is an important * Test preparation method* to follow

**even in practice test**environment

*as metrics or performance measurement is built-in.*

**for continuous skill-set improvement,**### 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 from the companion solution set 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, 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.

If you have not taken the corresponding test yet, first take the test by referring to * SSC CGL level Question Set 56 on Trigonometry 5* in prescribed time and then only return to these solutions for gaining maximum benefits.

You may watch the **video solutions** of the questions in the **two-part** **video** below.

**Part 1: Q1 to Q5**

**Part 2: Q6 to Q10**

### 56th solution set- 10 problems for SSC CGL exam: 5th on Trigonometry - testing time 12 mins

**Problem 1.**

The maximum value of $(2\sin \theta + 3\cos\theta)$ is,

- $1$
- $2$
- $\sqrt{13}$
- $\sqrt{15}$

**Solution - Problem analysis and execution**

To find the maximum value, the given expression is converted to an expression involving a single trigonometric function using the compound angle relation of,

$c\sin (\theta +\alpha)=c\left(\sin {\theta}\cos \alpha + cos {\theta}\sin \alpha\right)$.

The coefficients $a$ and $b$ are substituted with,

$a=c\cos \alpha$, $b=c\sin \alpha$.

The value of the new coefficient $c$, is then,

$c=\sqrt{a^2+b^2}$.

It is in fact the hypotenuse of a right triangle in which $a$ and $b$ are the other two sides and $\angle \alpha$ is the base angle.

As maximum value of $\sin (\theta +\alpha)$ is 1, the maximum value of the given expression turns out to be,

$c=\sqrt{a^2+b^2}$.

In this case, $a=2$ and $b=3$, and so the desired maximum value is,

$\sqrt{2^2+3^2}=\sqrt{13}$.

**Answer:** c: $\sqrt{13}$.

**Key concepts and techniques used:** * Conversion to single term function* --

*--*

**maxima minima of trigonometric expressions***--*

**substitution technique***.*

**rich trigonometry concepts****Note:** For detailed treatment on maxima minima of trigonometric expressions, refer to the tutorial session on **Trigonometry concepts part 3, maxima (or minima) of trigonometric expressions.**

**Problem 2.**

Find the minimum value of $9\tan^2 \theta + 4\cot^2 \theta$,

- $15$
- $6$
- $9$
- $12$

**Solution - Problem analysis**

As $\tan$ or $\cot$ functions both have undefined maximum value, the given expression can have only a minimum value.

How to decide which method of *maxima minima finding technique* to be followed?

In this case the two trigonometric functions when multiplied with each other* cancel out leaving a purely numeric result* as these two form a pair of

*. This is the*

**inverse trigonometric functions as well as their powers are same***for applying the well known*

**guiding criteria***.*

**AM GM inequality technique**The AM (Arithmetic Mean) GM (Geometric Mean) inequality technique simply says,

For a set of values,

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

For a two term expression, $9\tan^2 \theta + 4\cot^2 \theta$, the AM of the two terms taken individually as two values, is,

$\text{AM} = \displaystyle\frac{9\tan^2 \theta + 4\cot^2 \theta}{2}$,

and the GM of the two terms is,

$\text{GM}=\sqrt{\left(9\tan^2 \theta)(4\cot^2 \theta\right)}=6$.

So applying the inequality condition,

$\displaystyle\frac{9\tan^2 \theta + 4\cot^2 \theta}{2} \geq 6$,

Or, $9\tan^2 \theta + 4\cot^2 \theta \geq 12$.

The given expression can only then be greater than or equal to 12. So its minimum value is 12.

**Answer:** d: $12$.

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

*--*

**Problem solving strategy***.*

**Maxima minima of trigonometric expressions -- AM GM inequality -- inverse trigonometric functions****Problem 3.**

If $\text{cosec} \theta -\cot \theta= \displaystyle\frac{7}{2}$, the value of $\text{cosec} \theta$ is,

- $\displaystyle\frac{49}{28}$
- $\displaystyle\frac{53}{28}$
- $\displaystyle\frac{47}{28}$
- $\displaystyle\frac{51}{28}$

**Solution - Problem analysis and solving**

From friendly function pair concept we know,

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

So from the given value, $\text{cosec} \theta -\cot \theta= \displaystyle\frac{7}{2}$, we have,

$\text{cosec} \theta +\cot \theta= \displaystyle\frac{2}{7}$.

Adding the two equations,

$2\text{cosec} \theta = \displaystyle\frac{2}{7}+\displaystyle\frac{7}{2}$

$=\displaystyle\frac{53}{14}$,

Or, $\text{cosec} \theta = \displaystyle\frac{53}{28}$.

**Answer:** b: $\displaystyle\frac{53}{28}$.

**Key concepts and techniques used:** * Friendly function pairs concept* --

*Basic algebra concepts*

**.**

**Problem 4.**

If $\tan^2 \alpha=1 +2\tan^2 \beta$, where $\alpha$ and $\beta$ both are positive acute angles, find the value of $\sqrt{2}\cos \alpha - \cos \beta$.

- $0$
- $\sqrt{2}$
- $-1$
- $1$

**Solution - Problem analysis and solving**

The single given equation must be modified to produce the result. How to modify it?

Examining the equation, it seems to be unbalanced. The shortage of 1 on the RHS as well as on the LHS became visible in no time, because whenever there are two $\tan^2$s in the input and two $\cos$'s in the output, the relation of $sec^2$ with $\tan^2$ comes to mind.

So we add 1 to both sides, getting,

$1+\tan^2 \alpha=2 +2\tan^2 \beta$,

Or, $\sec^2 \alpha=2\sec^2 \beta$,

Or, $2\cos^2 \alpha = cos^2 \beta$.

So,

$\sqrt{2}\cos \alpha = \cos \beta$,

Or, $\sqrt{2}\cos \alpha - \cos \beta=0$.

**Answer:** a: $0$.

**Key concepts and techniques used:** * Expression balance assessment* --

*-*

**Missing element introduction technique***.*

**End state analysis approach****Problem 5.**

The value of $152(\sin 30^0 + 2\cos^2 45^0 + 3\sin 30^0 +$

$\hspace{30mm}4\cos^2 45^0 + ....+17\sin 30^0+18\cos^2 45^0)$ is,

- an irrational number
- a rational number but not an integer
- an integer but not a perfect square
- the perfect square of an integer

**Solution - Problem analysis**

By a quick look at the terms we find,

$\sin 30^0=\frac{1}{2}$ in every odd term and also,

$\cos^2 45^0=\frac{1}{2}$ in every even term.

Then $\frac{1}{2}$ can be taken out of the brackets as a common factor thus dividing 152 by 2 with a result of 76 and leaving a sum of natural numbers inside the brackets,

$1+2+3+.....+17+18$.

This is nothing but the sum of first 18 natural numbers which we evaluate quickly to be $9\times{17}+18=9\times{19}$,

Mechanism:$9$ is the middle term of the first 17 natural numbers andis the average of these 17 numbers, we add another 18 to get the sum of first 18 natural numbers.

Finally then the given equation evaluates to,

$76\times{9}\times{19}=36\times{19}\times{19}$, a perfect square of an integer, $6\times{19}=114$.

**Answer:** d: the perfect square of an integer.

**Key concepts and techniques used:** * Common pattern identification -- Problem transformation* to finding sum of first 18 natural numbers

**-- Efficient simplification, we didn't carry out the multiplication but kept the factors visible for later use for finding the nature of the number -- basic factors and multiples concept-- basic trigonometry concepts.**

**Problem 6.**

If $\tan \theta + \cot \theta=2$, then the value of $\tan^n \theta + \cot^n \theta$ ($0^0 \lt \theta \lt 90^0$, and $n$ an integer) is,

- $2$
- $2^{n+1}$
- $2^n$
- $2n$

**Solution - **Problem analysis and solving

As the functions are directly raised to $n$th power without any extra help, each of the functions must be 1. Let us verify our hypothesis.

The given equation,

$\tan \theta + \cot \theta=2$,

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

Or, $\tan^2 \theta -2\tan \theta + 1=0$,

Or, ,$(\tan \theta - 1)^2=0$,

Or, $\tan \theta = \cot \theta =1$.

So,

$\tan^n \theta + \cot^n \theta=2$.

**Answer:** a: $2$.

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

*--*

**Input transformation***.*

**basic trigonometric concepts****Problem 7.**

If $\displaystyle\frac{\sin \theta}{1+\cos \theta} + \displaystyle\frac{\sin \theta}{1-\cos \theta} = 4$, the value of $\cot \theta + \sec \theta$ is,

- $\sqrt{3}$
- $\displaystyle\frac{\sqrt{3}}{7}$
- $\displaystyle\frac{\sqrt{3}}{5}$
- $\displaystyle\frac{5}{\sqrt{3}}$

**Solution - problem analysis and solving**

We visualize the possibility of eliminating the denominators of each term separately and then form a simple addition,

$\displaystyle\frac{\sin \theta}{1+\cos \theta} + \displaystyle\frac{\sin \theta}{1-\cos \theta} = 4$,

Or, $\displaystyle\frac{1}{\text{cosec} \theta+\cot \theta} + \displaystyle\frac{1}{\text{cosec} \theta-\cot \theta}=4$, we bring down the $\sin \theta$ from the numerator to the denominator by dividing it

Or, $(\text{cosec} \theta - \cot \theta) + (\text{cosec} \theta + \cot \theta)=4$, using friendly function pair concept,

Or, $2\text{cosec} \theta = 4$,

Or, $\sin \theta = \frac{1}{2}$, that is, $\theta=30^0$.

So,

$\cot \theta = \cot 30^0=\sqrt{3}$, and,

$\sec \theta = \sec 30^0=\displaystyle\frac{2}{\sqrt{3}}$.

Thus we have the target expression,

$\cot \theta + \sec \theta=\sqrt{3}+\displaystyle\frac{2}{\sqrt{3}}=\displaystyle\frac{5}{\sqrt{3}}$

**Answer:** d: $\displaystyle\frac{5}{\sqrt{3}}$.

**Key concepts used:** * Key pattern identification* --

*--*

**Denominator elimination***--*

**Friendly trigonometric function pair**

**Basic trigonometry concepts.****Problem 8.**

If $\sin \theta + \cos \theta =\sqrt{2}$ with $\angle \theta$ a positive acute angle, then the value of $\tan \theta + \sec \theta$ is,

- $\sqrt{3}-1$
- $\displaystyle\frac{1}{\sqrt{2}-1}$
- $\sqrt{2}-1$
- $\sqrt{3}+1$

**Solution - problem analysis and solving**

The given equation is,

$\sin \theta + \cos \theta =\sqrt{2}$

Squaring both sides,

$\sin^2 \theta + 2\sin {\theta}\cos \theta + \cos^2 \theta=2$,

Or, $2\sin {\theta}\cos \theta = 1$,

Or, $\sin^2 \theta + \cos^2 \theta - 2\sin {\theta}\cos \theta=1-1=0$,

Or, $(\sin \theta - \cos \theta)^2=0$,

Or, $\sin \theta - \cos \theta = 0$,

Or, $\sin \theta=\cos \theta=\displaystyle\frac{1}{\sqrt{2}}$, from $\sin \theta + \cos \theta = \sqrt{2}$.

Thus,

$\tan \theta=1$, and,

$\sec \theta=\sqrt{2}$.

So,

$\tan \theta + \sec \theta = \sqrt{2}+1$.

Multiplying the numerator and denominator of RHS by $(\sqrt{2}-1)$,

$\tan \theta +\sec \theta =\displaystyle\frac{1}{\sqrt{2}-1}$.

**Answer:** b: $\displaystyle\frac{1}{\sqrt{2}-1}$.

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

*--*

**basic trigonometric concepts***.*

**surd rationalization -- basic algebra concepts****Note;** Without any deduction also, from the given expression, value of $\theta=45^0$ can be guesssed by testing the given expression.

**Problem 9.**

If $p=a\sec {\theta}\cos \alpha$, $q =b\sec {\theta}\sin \alpha$, and $r =c\tan \theta$, then the value of $\displaystyle\frac{p^2}{a^2} +\displaystyle\frac{q^2}{b^2}-\displaystyle\frac{r^2}{c^2}$ is,

- 0
- 1
- 4
- 5

**Solution - problem analysis and solving**

Analyzing the target expression with respect to the given expression we decide to form the three individual terms of the target expression from the three input expressions by input transformation.

The first given equation is,

$p=a\sec {\theta}\cos \alpha$,

Or, $\displaystyle\frac{p^2}{a^2}=\sec^2 {\theta}\cos^2 \alpha$.

Similarly from the second and third given expressions we get,

$\displaystyle\frac{q^2}{b^2}=\sec^2 {\theta}\sin^2 \alpha$, and

$\displaystyle\frac{r^2}{c^2}=\tan^2 {\theta}$.

Adding the first two transformed expressions,

$\displaystyle\frac{p^2}{a^2}+ \displaystyle\frac{q^2}{b^2}=\sec^2 {\theta}\cos^2 \alpha +\sec^2 {\theta}\sin^2 \alpha$,

Or, $\displaystyle\frac{p^2}{a^2}+ \displaystyle\frac{q^2}{b^2}=\sec^2 \theta$

$=1+\tan^2 \theta$

$=1+\displaystyle\frac{r^2}{c^2}$

So,

$\displaystyle\frac{p^2}{a^2} +\displaystyle\frac{q^2}{b^2}-\displaystyle\frac{r^2}{c^2}=1$

**Answer:** b: 1.

**Key concepts and techniques used:** * End state analysis comparing target with givens comparing similarities* --

*--*

**input transformation**

**basic trigonometric concepts***.*

**-- basic algebra concepts****Problem 10.**

$\displaystyle\frac{\sin^2 \theta}{\cos^2 \theta}+\displaystyle\frac{\cos^2 \theta}{\sin^2 \theta}$ is equal to,

- $\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta}$
- $\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta} -2$
- $\displaystyle\frac{1}{\tan^2 \theta - \cot^2 \theta}$
- $\displaystyle\frac{\sin^2 \theta}{\cot \theta - \sec \theta}$

#### Solution - Problem analysis

As combining the two terms results in $\sin^4 \theta+\cos^4 \theta$ as numerator, this option seems to lead to a more complex solution path and so is rejected.

On brief examination of the choice values, the $-2$ term in the Option 2 attracts attention and prompts us more towards adding and subtracting 2 to the expression, and breaking down the problem into **individual term simplification. **

The 2 added is allocated 1 to each term to eliminate the numerator and simplify the numerator considerably.

Instead of Denominator elimination technique here we eliminate the numerator to the value of 1 by adding 1 to each term.

#### Solution - Problem solving execution

Adding and subtracting 2 from the expression and allocating 1 to each term for addition we obtain,

$\displaystyle\frac{\sin^2 \theta}{\cos^2 \theta}+\displaystyle\frac{\cos^2 \theta}{\sin^2 \theta}$

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

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

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

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

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

**Answer:** b: $\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta}-2$.

**Key concepts and techniques used:** * Problem analysis*, combining the the two terms seems to lead to a more complex path, and so is rejected --

*, as the term $-2 $ in choice 2 prompted us to add 2 and subtract 2 --*

**End state analysis***, individual terms are simplified first --*

**Problem breakdown technique***, often simplification of each term in an expression speeds up the solution considerably --*

**Individual term simplification***, the two 1s added to the two terms to simplify each considerably --*

**Introduction of new simplifying terms***, the numerator of each term is converted to 1, effectively eliminating it --*

**Numerator elimination***--*

**Basic trigonometry concepts***--*

**Basic algebra concepts***.*

**Rich algebra concepts and techniques**

**Note:** You will observe that in many of the Trigonometric problems basic and 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

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**How to solve a difficult SSC CGL level problem in a few reasoned steps, Trigonometry 10**

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**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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