## 1st Preparatory SSC CGL level Solution Set, topic Trigonometry 1

This is the 1st solution set for the 10 preparatory level practice problem exercise for SSC CGL exam and 1st on topic Trigonometry. You may refer to the * 1st Preparatory SSC CGL level questions with answers set and 1st 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,**you may refer to our tutorialor any short but good material to refresh your concepts if you so require.**Basic and rich Trigonometric concepts and applications****Answer the questions**in an undisturbed environment with no interruption, full concentration and alarm set at 15 minutes.**When the time limit of 15 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 15 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**

**You may refer to:**

* 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.**

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### 1st solution set- 10 preparatory level problems for SSC CGL exam: 1st on Trigonometry - testing time 15 mins

**Problem 1.**

If $sin (A+B)=sin Acos B + cos Asin B$, then the value of $\sin 75^0$ is,

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

**Solution 1 - Problem analysis**

Compound angle trigonometric knowledge is not wanted, rather, a compound angle relationship has been given for $sin(A+B)$.

As all the terms on the RHS are in terms of $sin A$, $sin B$, $cos A$ and $cos B$, we must select $A$ and $B$ in such a way that, $A+B=75^0$ and also the trigonometric ratio term values on the RHS are commonly used and known.

This is the main challenge for solving the problem—finding a suitable combination of $A$ and $B$.

**Solution 1 - Problem solving execution**

Soon as the challenge was known, we could identify the best combination as,

$A + B=45^0+30^0=75^0$.

Substituting the related values in RHS,

$sin (A+B)=\displaystyle\frac{1}{\sqrt{2}}\times{\displaystyle\frac{\sqrt{3}}{2}} + \displaystyle\frac{1}{\sqrt{2}}\times{\displaystyle\frac{1}{2}}$

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

**Answer:** Option d: $\displaystyle\frac{\sqrt{3}+1}{2\sqrt{2}}$.

**Key concepts and techniques used:** * Commonly used values of trigonometric ratios* -- basic trigonometry concepts.

**Problem 2.**

If the $A$, $B$, $C$ and $D$ are the angles of a cyclic quadrilateral, the value of $cos A + cos B + cos C + cos D$ is,

- $2$
- $0$
- $1$
- $-1$

**Solution 2 - Problem analysis**

For a * cyclic quadrilateral* ABCD, opposite angles sum up to $180^0$ so that,

$cos A = cos(180^0- C)=-cos C$, and

$cos B = cos(180^0- D)=-cos D$.

So,

$cos A + cos B + cos C + cos D=0$.

We have used the relation between opposite angles of a cyclic quadrilateral and also the * supplementary trigonometric function* concepts.

**Answer:** Option b: $0$.

**Key concepts and techniques used**: * Cyclic quadrilateral -- Supplementary trigonometric functions*.

#### Supplementary angle trigonometric functions

Two trigonometric functions are supplementary functions if, the sum of the two angles $A$ and $C$ is $180^0$.

Some of the supplemetary angle relationships,

$cos A = cos(180^0- C)=-cos C$,

$sin A = sin(180^0- D)=sin C$,

$tan A = tan(180^0- D)=-tan C$.

**Problem 3.**

If $cos^2 \alpha - sin^2 \alpha = tan^2 \beta$, then the value of $cos^2 \beta - sin^2 \beta$ is,

- $cot^2 \beta$
- $tan^2 \alpha$
- $cot^2 \alpha$
- $tan^2 \beta$

**Solution 3 - Problem analysis**

From the target expression we find, we need to evaluate $cos^2 \beta$ and $sin^2 \beta$ in terms of functions in $\alpha$ from the given equation.

By * trigonometric function derivation principle*, it is always possible to transform square of any trigonometric function to square of any other trigonomtric function.

**Solution 3 - Problem solving execution**

Given equation,

$cos^2 \alpha - sin^2 \alpha = tan^2 \beta$

Or, $cos^2 \alpha - 1+ cos^2 \alpha = sec^2 \beta - 1$

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

Or, $cos^2 \beta=\displaystyle\frac{1}{2}sec^2 \alpha$.

Or, $sin^2 \beta=1 -\displaystyle\frac{1}{2}sec^2 \alpha$.

Subtracting,

$cos^2 \beta - sin^2 \beta=sec^2 \alpha -1=tan^2 \alpha$.

**Answer:** Option b: $tan^2 \alpha$.

**Key concepts and techniques used:** * Basic trigonometry concept* --

*.*

**Trigonometric function derivation principle****Problem 4.**

If $cos^4 \theta - sin^4 \theta=\displaystyle\frac{2}{3}$, then the value of $1 - 2sin^2 \theta$ is,

- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{3}{2}$
- $0$
- $1$

**Solution 4 - Problem analysis and solving**

Using basic algebraic relation of $a^2 - b^2=(a+b)(a-b)$, we transfom the given equation to,

$cos^4 \theta - sin^4 \theta=\displaystyle\frac{2}{3}$

Or, $(cos^2 \theta + sin^2 \theta)(cos^2 \theta - sin^2 \theta)=\displaystyle\frac{2}{3}$

Or, $(cos^2 \theta - sin^2 \theta)=\displaystyle\frac{2}{3}$.

Or, $1 -sin^2 \theta - sin^2 \theta=\displaystyle\frac{2}{3}$,

Or, $1 -2sin^2 \theta=\displaystyle\frac{2}{3}$.

RHS remained unchanged and only LHS needed transformation.

**Answer:** Option a: $\displaystyle\frac{2}{3}$.

**Key concepts and techniques used: Basic trigonometry concepts** --

**Basic algebra concepts****.****Problem 5.**

If $x=asec \theta$ and $y=btan \theta$, then $\displaystyle\frac{x^2}{a^2}-\displaystyle\frac{y^2}{b^2}$ is,

- $0$
- $2$
- $1$
- $-1$

**Solution 5 - Problem solving execution**

From the given inputs we will straightway form the ratios of $x$ and $a$ and $y$ and $b$ and take a subtraction of quares as is wanted,

$\displaystyle\frac{x^2}{a^2}-\displaystyle\frac{y^2}{b^2}=sec^2 \theta - tan^2 \theta=1$

**Answer:** Option c: $1$.

We used the basic relation between friendly trigonometric function pair, $sec \theta$ and $tan \theta$ and target driven algebraic simplification.

**Problem 6.**

The value of $\displaystyle\frac{\sin \theta - 2sin^3 \theta}{2cos^3 \theta - cos \theta}$ is,

- $sin \theta$
- $cot \theta$
- $cos \theta$
- $tan \theta$

**Solution 6 - **Problem analysis

$sin \theta$ and $cos \theta$ appear both in numerator and denominator in similar forms and ready to be factored out,

$\displaystyle\frac{\sin \theta - 2sin^3 \theta}{2cos^3 \theta - cos \theta}$

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

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

$=tan \theta$

**Answer:** Option d: $tan \theta$.

**Key concepts and techniques used:** * Basic trigonometry copncepts *--

*Factoring out technique -- Use of friendly trigonometric function pair, $sin \theta$ and $cos \theta$.***Problem 7.**

If $sin \theta + cos \theta=p$ and $sec \theta + cosec \theta=q$, then the value of $q(p^2 - 1)$ is,

- $2p$
- $p$
- $2$
- $1$

**Solution 7 - Problem solving execution **

We will straightaway start deriving the value of target expression with larger factor first,

$p^2-1$,

$=(sin \theta +cos \theta)^2 - 1$

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

So,

$q(p^2-1)$

$=(2sin \theta.cos \theta)(sec \theta + cosec \theta)$

$=2(sin \theta + cos \theta)$

$=2p$.

**Answer:** Option a: $2p$.

**Key concepts used:** * Basic trigonometry concepts* --

*--*

**Basic algebra concepts***--*

**Pattern identification technique***.*

**Substitution****Problem 8.**

If $sin (3\alpha - \beta)=1$, and $\cos (2\alpha + \beta)=\displaystyle\frac{1}{2}$, then the value of $tan \alpha$ is,

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

**Solution 8 - Problem solving execution**

We will directly evaluate the bracketed angles from given trigonometric term ratio values,

$sin (3\alpha - \beta)=1$.

So,

$3\alpha - \beta=90^0$.

Again,

$cos (2\alpha + \beta)=\displaystyle\frac{1}{2}$

So,

$2\alpha + \beta=60^0$.

Adding the two,

$5 \alpha=150^0$,

Or, $\alpha=30^0$,

Or, $tan \alpha=tan 30^0 =\displaystyle\frac{1}{\sqrt{3}}$.

**Answer:** Option b: $\displaystyle\frac{1}{\sqrt{3}}$.

**Key concepts and techniques used:** **Basic trigonometry concepts -- Trigonometric term values.**

**Problem 9.**

If $rsin \theta =\displaystyle\frac{7}{2}$ and $rcos \theta =\displaystyle\frac{7\sqrt{3}}{2}$, then the value of $r$ is,

- $3$
- $4$
- $5$
- $7$

**Solution** 9 - Problem solving execution

We straightaway eliminate both $sin \theta$ and $cos \theta$ by first isolation (dividing by $r$), squaring and then adding,

$sin^2 \theta + cos^2 \theta = \displaystyle\frac{49}{4r^2}+\displaystyle\frac{3\times{49}}{4r^2}$

Or, $r^2=49$,

Or, $r=7$.

**Answer:** Option d: $7$.

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

*--*

**Basic trigonometry concepts***.*

**efficient simplification**All in mind and in quick time.

**Problem 10.**

If $0^0 \lt \theta \lt 90^0$ and $2sin^2 \theta + 3cos \theta=3$, then the value of $\theta$ is,

- $30^0$
- $45^0$
- $60^0$
- $75^0$

#### Solution 10 - Problem analysis

The equation involves $cos \theta$ and $sin^2 \theta$. If only we convert $sin^2 \theta$ to $1-cos^2 \theta$, we would get a quadratic equation in $cos \theta$.

Out of the two possible roots of $cos \theta$ we expect one to be eliminated by the first range condition on $\theta$.

#### Solution 10. Problem solving execution

Given equation,

$2sin^2 \theta + 3cos \theta=3$

Or, $2 - 2cos^2 \theta + 3cos \theta=3$,

Or, $2cos^2 \theta - 3cos \theta + 1=0$.

Or, $(2cos \theta - 1)(cos \theta - 1)=0$.

But as, $ 0^0 \lt \theta \lt 90^0$, $cos \theta \neq 1$ leaving, $2cos \theta = 1$, or, $\theta=60^0$.

**Answer:** Option c: $60^0$.

**Key concepts and techniques used:** *Basic trigonometry concepts -- basic algebraic concepts -- roots of quadratic equation -- trigonometric ratio term values.*

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

### SSC CHSL question and solution sets on Trigonometry

**Preparatory SSC CGL level Solutions on questions set 1 Trigonometry 1**

**Preparatory SSC CGL level Questions with answers set 1 Trigonometry 1**

### Tutorials on Trigonometry

**Basic and rich concepts in Trigonometry and its applications**

### General guidelines for success in SSC CGL and similar exams

**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.

All Trigonometry related materials are avalable on * Suresolv Trigonometry*. A large part of these are on SSC CGL.

### 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**