## Sixteenth SSC CGL level Solution Set, topic Trigonometry

This is the sixteenth solution set of 10 practice problem exercise for SSC CGL exam on topic Trigonometry. Students should complete the corresponding question set in prescribed time first and then only refer to the solution set.

We found from our analysis of the Trigonometry problems that this topic is built on a small set of basic and rich concepts. It makes possible for you to solve any problem in this topic area fast and quick, following elegant problem solving methods.

We have tried to show you how this can be done in this solution set. But before going through these solutions please take the test first in prescribed time by referring to * SSC CGL level Question Set 16 on Trigonometry*.

Watch the **video solutions** in the **two-part video**.

**Part 1: Q1 to Q5**

**Part 2: Q6 to Q10**

### Sixteenth solution set- 10 problems for SSC CGL exam: topic Trigonometry - time 12 mins

**Problem 1.**

The simplified value of $(sec\theta - cos\theta)^2 + (cosec\theta - sin\theta)^2 - (cot\theta - tan\theta)^2$ is,

- $\displaystyle\frac{1}{2}$
- $0$
- $2$
- $1$

**Solution:**

**Conventional approach is to expand all the three squares,** transform and simplify the terms and get the simplified target value. But as is the norm, that is a longer time-consuming path. We have to search for fast elegant solution.

**First clue:**

The third square expression being subtracted, a prospective approach is to simplify the first two square expressions in hopefully terms of $tan\theta$ and $cot\theta$ so that these can be combined with the negative terms from the expanded third expression for achieving simplified final result.

Looking at the first two terms we identify a **rich concept** that will help us do that.

#### Rich concept of Factoring out the inverse

Being well-acquainted with basic trigonometric concepts, whenever we encounter an expression of $(sec\theta - cos\theta)$, a sum of inverses, we recognize that we can take the inverse $sec\theta$ out of the brackets leaving the well known friendly expression $(1 - cos^2\theta)$ within the brackets, that will immediately be transformed to $sin^2\theta$. Thus a **sum of terms will be simplified to a product of terms.**

In any expression manipulation, we always look for such opportunities to reduce the number of terms in expressions so that product of sum of terms expressions are transformed to product of terms. This is a general algebraic rich concept and technique for simplification.

$(sec\theta - cos\theta) = sec\theta(1 - cos^2\theta)$

$\hspace{10mm}= sec\theta{sin^2\theta} = sin\theta{tan\theta}$, and similarly,

$(cosec\theta - sin\theta) = cosec\theta(1 - sin^2\theta)$

$\hspace{10mm}= cosec\theta{cos^2\theta} = cos\theta{cot\theta}$.

These are very useful patterns for application of the **rich concept of factoring out the inverse.**

Let us focus on the first two terms then and apply the rich concept and technique of factoring out the inverse,

$(sec\theta - cos\theta)^2 = sec^2\theta(1 - cos^2\theta)^2 = {sin^2\theta}tan^2\theta$.

Similarly,

$(cosec\theta - sin\theta)^2 = {cos^2\theta}cot^2\theta$.

Now we sum up these terms with the expanded third term and apply **principle of collection of friendly terms. **The pair of terms involving $tan^2\theta$ and the second pair involving $cot^2\theta$ are combined together.

$E = 2 - tan^2\theta(1 - sin^2\theta) - cot^2\theta(1 - cos^2\theta)$

$\hspace{12mm}= 2 - (sin^2\theta + cos^2\theta)$

$\hspace{12mm}= 1$.

**Answer:** Option d: $1$.

**Key concepts used:** Transforming the first two expressions to single term expressions with $tan^2\theta$ and $cot^2\theta$ as factors with the intention of merging with the expanded corresponding terms from the third expression. This is achieved by applying the * rich concept of factoring out the inverse* $sec\theta$ and $cosec\theta$ -- use of

*to find similarities between the third expression terms and the first two expressions --*

**end state analysis***of $tan^2\theta$ and $cot^2\theta$ together to cancel out the denominators and leave only $sin^2\theta$ and $cos^2\theta$.*

**collecting the friendly terms****Problem 2.**

If $\displaystyle\frac{sin\theta + cos\theta}{sin\theta - cos\theta} = \frac{5}{4}$, then the value of $\displaystyle\frac{tan^2\theta + 1}{tan^2\theta - 1}$ will be,

- $\displaystyle\frac{41}{40}$
- $\displaystyle\frac{40}{41}$
- $\displaystyle\frac{25}{16}$
- $\displaystyle\frac{41}{9}$

**Solution:**

Applying the powerful **componendo dividendo technique** on the input expression we get directly,

$\displaystyle\frac{sin\theta}{cos\theta} = \displaystyle\frac{5 + 4}{5 - 4} = 9$,

Or, $tan\theta = 9$

Substituting in the target expression,

$E = \displaystyle\frac{9^2 + 1}{9^2 - 1} = \displaystyle\frac{82}{80} = \displaystyle\frac{41}{40}$

**Answer:** Option a : $\displaystyle\frac{41}{40}$.

**Key concepts used:** **Componendo dividendo** -- substitution.

#### Rich algebraic technique of Componendo and dividendo

**Problem:** Simplify, $\displaystyle\frac{x + y}{x - y} = 3$.

First we add 1 to both sides of the equation,

$\displaystyle\frac{x + y}{x - y} + 1 = 3 + 1$,

Or, $\displaystyle\frac{2x}{x - y} = 4$.

Next we subtract 1 from both sides of the equation,

$\displaystyle\frac{x + y}{x - y} - 1 = 3 - 1$,

Or, $\displaystyle\frac{2y}{x - y} = 2$.

Now we divide the first result by the second,

$\displaystyle\frac{x}{y} = 2$, a greatly simplified expression with two terms in the numerator and denominator reduced to single terms.

This is a powerful algebraic technique frequently applied whenever we encounter the special form of given expression.

**Problem 3.**

If $sin\theta + cosec\theta =2$, then the value of $sin^{100}\theta + cosec^{100}\theta$ is,

- 100
- 3
- 2
- 1

**Solution:**

Whenever the target expression involves large powers of $sin\theta$ or $cos\theta$, we can be sure that from the input expression we would get either the value of $\theta$ or the value of $sin\theta$, as otherwise such large powers can't be evaluated.

With this prior knowledge we proceed to expand the given expression,

$sin\theta + cosec\theta = 2$,

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

Or, $sin\theta = 1$, and so, $cosec\theta=1$.

Putting these convenient values in the target expression we get,

$E = 1 + 1 = 2$

**Answer:** Option c: 2.

**Key concepts used:** Deciding from target expression that value of $sin\theta$ must be obtained from the given expression -- expanding and evaluating the given expression produced the convenient value of $sin\theta$.

**Problem 4.**

The greatest value of $sin^4\theta + cos^4\theta$ is,

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

**Solution:**

Using algebraic maxima finding technique we transform part of the given expression to a square of sums.

$sin^4\theta + cos^4\theta $

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

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

The maximum of this expression can only be 1 when, the second term is zero, or when either $sin\theta$ or $cos\theta$ is 0.

This is a quick method as it has used the trigonometric relations along with the algebraic maxima technique elegantly.

**Answer:** Option a: $1$.

**Key concepts used:** To use the * maxima technique for quadratic equations*, converting part of the given expression to square of sums -- finding the maximum condition.

**Problem 5.**

If $\displaystyle\frac{sin\theta}{x} = \displaystyle\frac{cos\theta}{y}$, then $sin\theta - cos\theta$ is,

- $x - y$
- $\displaystyle\frac{x - y}{\sqrt{x^2 + y^2}}$
- $\displaystyle\frac{y - x}{\sqrt{x^2 + y^2}}$
- $x + y$

**Solution:**

Though the target expression looks simple, the task is cut out for us, we must express both $sin\theta$ and $cos\theta$ in terms of $x$ and $y$.

With this resolve we start manipulating the given expression in the simplest path possible,

$\displaystyle\frac{sin\theta}{x} = \displaystyle\frac{cos\theta}{y}$,

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

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

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

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

Similarly,

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

Thus target expression,

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

**Answer:** Option b: $\displaystyle\frac{x - y}{\sqrt{x^2 + y^2}}$.

**Key concepts used:** Using the rich concept of $cosec^2\theta = cot^2\theta + 1$ and $sec^2\theta = tan^2\theta + 1$ -- expressing $sin\theta$ and $cos\theta$ in terms of $x$ and $y$.

**Problem 6.**

If $tan\theta - cot\theta = 0$ find the value of $sin\theta + cos\theta$.

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

** Solution:**

With the form of the given expression we will go straight for equating the $tan$ and $cot$ with the intention of getting an equation in $sin$ and $cos$.

So,

$tan\theta = cot\theta$,

Or, $sin^2\theta = cos^2\theta$.

We know this equality occurs only when $sin\theta = cos\theta = \displaystyle\frac{1}{\sqrt{2}}$ at $\theta = 45^0$.

So, $sin\theta + cos\theta = 2\times{\displaystyle\frac{1}{\sqrt{2}}} = \sqrt{2}$

**Answer:** Option a : $\sqrt{2}$.

**Key concepts used:** Transforming the given expression in terms of $sin$ and $cos$ to get the equality condition.

**Problem 7.**

If $sin21^0 = \displaystyle\frac{x}{y}$ then $sec21^0 - sin69^0$ is,

- $\displaystyle\frac{y^2}{x\sqrt{y^2 - x^2}}$
- $\displaystyle\frac{x^2}{y\sqrt{y^2 - x^2}}$
- $\displaystyle\frac{x^2}{y\sqrt{x^2 - y^2}}$
- $\displaystyle\frac{y^2}{x\sqrt{x^2 - y^2}}$

** Solution:**

We know from the beginning that we have to use the rich trigonometric concept of, $sin\theta = cos\left({\displaystyle\frac{\pi}{2} - \theta}\right)$.

In our problem situation applying this concept we have, $sin21^0 = cos 69^0$,

So, $sin21^0 = cos69^0 = \displaystyle\frac{x}{y}$,

Or, $1 - cos^2{69^0} = sin^2{69^0} = \displaystyle\frac{y^2 - x^2}{y^2}$,

Or, $sin{69^0} = \displaystyle\frac{\sqrt{y^2 - x^2}}{y}$.

The target expression,

$sec21^0 - sin69^0 = cosec69^0 - sin69^0$

$= \displaystyle\frac{1 - sin^2{69^0}}{sin69^0}$

$= \displaystyle\frac{cos^2{69^0}}{sin69^0}$,

$=\displaystyle\frac{x^2}{y^2}\times{\displaystyle\frac{y}{\sqrt{y^2 - x^2}}}$

$=\displaystyle\frac{x^2}{y\sqrt{y^2 - x^2}}$.

**Answer:** Option b: $\displaystyle\frac{x^2}{y\sqrt{y^2 - x^2}}$.

**Key concepts used:** Rich concept of $sin\theta = cos\left({\displaystyle\frac{\pi}{2} - \theta}\right)$ to get $cos69^0$ and then $sin69^0$ -- * trigonometric simplification first* then only substituting the complex value at the end to minimize calculations.

**Problem 8.**

If $\displaystyle\frac{sec\theta+ tan\theta}{sec\theta - tan\theta}=\displaystyle\frac{5}{3}$ then $sin\theta$ is,

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

**Solution:**

As usual detecting the possibility of taking out $sec\theta$ as a factor from both numerator and denominator we take up the transformation,

$\displaystyle\frac{sec\theta+ tan\theta}{sec\theta - tan\theta}=\displaystyle\frac{5}{3}$,

Or, $\displaystyle\frac{sec\theta(1 + sin\theta)}{sec\theta(1 - sin\theta)}=\displaystyle\frac{5}{3}$,

Or, $\displaystyle\frac{1 + sin\theta}{1 - sin\theta}=\displaystyle\frac{5}{3}$.

This is ripe form for applying the Componendo dividendo technique. We thus get,

$sin\theta = \displaystyle\frac{5 - 3}{5 + 3} = \displaystyle\frac{1}{4}$

**Answer:** Option d: $\displaystyle\frac{1}{4}$.

**Key concepts used:** Recognition that $sec\theta$ can be * factored out *from both numerator and denominator -- and then use

**componendo dividendo technique.****Problem 9.**

If $(1 + sin A)(1 + sin B)(1 + sin C) = (1 - sin A)(1 - sin B)( 1 - sin C)$, then the expression on each side of the equation equals,

- $1$
- $tan A.tan B.tan C$
- $cos A.cos B.cos C$
- $sin A.sin B.sin C$

**Solution: **

Observing the possibility of getting $1 - cos^2\theta$ for each of the three expressions on one side if we multiply them together we first equate the expressions to a dummy variable $p$,

$(1 + sin A)(1 + sin B)(1 + sin C) $

$\hspace{5mm}= (1 - sin A)(1 - sin B)( 1 - sin C) = p$ which results in two equations,

$(1 + sin A)(1 + sin B)(1 + sin C)= p$ and,

$(1 - sin A)(1 - sin B)( 1 - sin C) = p$.

Multiplying the two together,

$p^2 = (1 - sin^2 A)(1 - sin^2 B)(1 - sin^2 C)$

$=cos^2 A. cos^2 B.cos^2 C$.

Or, each target expression $p=cos A.cos B.cos C$.

**Answer:** Option c: $cos A.cos B.cos C$.

**Key concepts used:** Identifying the possibility of getting simplified $1 - cos^2\theta$ for each of the three sum terms if multipled together we introduce a dummy variable equaling the expressions -- multiplication and simplification. This is simple algebraic technique using basic trigonometric concepts.

**Problem 10.**

If $\theta = 60^0$, then, $\displaystyle\frac{1}{2}\sqrt{1 + sin\theta} + \displaystyle\frac{1}{2}\sqrt{1 - sin\theta}$ is,

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

**Solution:**

Substituting value of $\theta$,

$\sqrt{1 + sin\theta}$

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

$= \sqrt{\displaystyle\frac{2 + \sqrt{3}}{2}}$

The problem expression is thus $\sqrt{1 + sin\theta}$ in which we need to eliminate the square root applying the powerful surd techniques of transforming a two term surd expression to a square of sum of a two term surd expression.

As the surd $\sqrt{3}$ doesn't have 2 as a coefficient, we multiply and diivide by 2, getting,

$\sqrt{1 + sin\theta}$

$= \sqrt{\displaystyle\frac{(\sqrt{3} + 1)^2}{4}}$

$=\displaystyle\frac{\sqrt{3} + 1}{2}$,

Similarly,

$\sqrt{1 - sin\theta} = \displaystyle\frac{\sqrt{3} - 1}{2}$, and the target expression,

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

**Answer:** Option a: $cos\displaystyle\frac{\theta}{2}$.

**Key concepts used:** Application of surd technique of transforming a two term surd expression to a square of a two term surd expression -- simplification.

**Note:** You will observe that in many of these Trigonometric problems rich algebraic concepts and techniques have been used. In fact that is the norm. Algebraic concepts are frequently used for elegant solutions of Trigonometric problems. But compared to difficulties of purely algebraic problem solving, trigonometry problems are simpler because by applying a few basic and rich trigonometric concepts along with algebraic concepts elegant solutions are reached faster.

### 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.

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**SSC CGL Tier II level Question set 12 Trigonometry 3, questions with answers**

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**SSC CGL Tier II level Question set 11 Trigonometry 2**

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**SSC CGL level Solution Set 19 on Trigonometry**

**SSC CGL level Question set 19 on Trigonometry**

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#### Algebraic concepts

**Basic and rich Algebraic concepts for elegant solutions of SSC CGL problems**

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