## 7th SSC CGL Tier II level Solution Set, topic Trigonometry 1

This is the 7th solution set for the 10 practice problem exercise for SSC CGL exam and 1st on topic Trigonometry.

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 through**Tutorial on Basic and rich concepts in Trigonometry and its applications,**or any short but good material to refresh your concepts if you so require.**Tutorial on Basic and rich concepts on Trigonometry part 2, compound angle functions****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**

#### 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 on Trigonometry part 2, compound angle functions.**

**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 Tier II level Question Set 7 on Trigonometry 1 * in prescribed time and then only return to these solutions for gaining maximum benefits.

If you wish, you may watch the two-part video solutions below.

**Part 1: Q1 to Q5**

**Part 2: Q6 to Q10**

### 7th solution set- 10 problems for SSC CGL Tier II exam: 1st on Trigonometry - testing time 15 mins

**Problem 1.**

If $x=rsin\alpha {cos\beta}$, $y=rsin\alpha{sin\beta}$ and $z=rcos\alpha$, then,

- $x^2-y^2+z^2=r^2$
- $x^2+y^2+z^2=r^2$
- $x^2+y^2-z^2=r^2$
- $y^2+z^2-x^2=r^2$

**Solution 1 - Problem analysis**

As we see it solving of the problem involves elimination of two classes of trigonometric functions in $\angle \alpha$ and $\angle \beta$. As we know, if any of the six trigonometric functions is available in a specific $\angle \alpha$ any other function in its class can be derived. That is the reason we talk about two classes of functions, though we have here four functions, $sin\alpha$, $sin\beta$, $cos\alpha$ and $\cos\beta$.

So finally the problem solving comes down to how quickly we can eliminate all of the trigonometric functions using algebraic concepts and also the * friendly trigonometric function pair concepts*, explained in more details in the solution of problem 5.

One of the standard approaches in such simplification cases is to get an expression in one variable fastest. This we can get simply by taking the ratio between the first two equations, getting,

$x=ycot\beta$.

Now we will derive $sin\beta$ from this result and substituting the value in the second equation we will get the value of $sin\alpha$.

The third equation also gives the value of $sin\alpha$. Equating these two expression of $sin\alpha$ we will be able to eliminate all of the trigonometric functions.

**Solution 1 - Problem solving execution**

$x=ycot\beta$,

Or, $cot^2\beta=\displaystyle\frac{x^2}{y^2}$

Or, $cosec^2\beta= \displaystyle\frac{x^2+y^2}{y^2}$

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

The second equation is,

$y=rsin\alpha{sin\beta}$,

Or, $y^2=r^2sin^2\alpha{sin^2\beta}=\displaystyle\frac{r^2y^2sin^2\alpha}{x^2+y^2}$,

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

Now is the time to turn our attention to the third equation,

$z=rcos\alpha$,

Or, $z^2=r^2\cos^2\alpha=r^2(1-sin^2\alpha)$,

Or, $x^2+y^2+z^2=r^2$.

**Answer:** b: $x^2+y^2+z^2=r^2$.

**Key concepts and techniques used:** Systematic elimination of multiplicative variables between three equations -- basic algebraic concepts -- basic trigonometry concepts -- * concepts of friendly trigonometric function pairs* -- efficient simplification --

**substitution technique.****Problem 2.**

With $\angle \theta$ acute, the value of the expression, $\left(\displaystyle\frac{5\ cos \theta - 4}{3-5\ sin \theta} - \displaystyle\frac{3+5\ sin \theta}{4+5\ cos \theta}\right)$ is,

- $1$
- $0$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{1}{4}$

**Solution 2 - Problem analysis**

Examining the numerator and denominator of the two fractions in the given expression, we identify the opportunity to use one of the most used * algebraic expression simplying resource* in the product of numerator of one with the denominator of the other as,

$(a+b)(a-b)=a^2-b^2$.

The individual expressions being quite assymmetric with no identifiable useful pattern, this should be the approach to be adopted.

#### Solution 2 - Problem solving execution

The given target expression,

$E=\left(\displaystyle\frac{5\ cos \theta - 4}{3-5\ sin \theta} - \displaystyle\frac{3+5\ sin \theta}{4+5\ cos \theta}\right)$,

$=\displaystyle\frac{\left((5cos \theta)^2 - 4^2\right)-\left(3^2-(5sin \theta)^2\right)}{(3-5\ sin \theta)(4+5\ cos \theta)}$,

$=\displaystyle\frac{25(sin^2 \theta +cos^2 \theta)-25}{(3-5\ sin \theta)(4+5\ cos \theta)}$,

$=0$.

**Answer:** b: 0.

**Key concepts and techniques used:** Key pattern identification -- * algebraic expression simplifying resource* -- basic algebra concepts --

**efficient simplification.****Problem 3.**

If $4+ 3\ tan \alpha=0$, where $\displaystyle\frac{\pi}{2} \lt \alpha \lt \pi$, the value of $2\ cot \alpha - 5\ cos \alpha + \sin \alpha$ is,

- $\displaystyle\frac{23}{10}$
- $-\displaystyle\frac{53}{10}$
- $\displaystyle\frac{37}{10}$
- $\displaystyle\frac{7}{10}$

**Solution 3 - Problem analysis**

The presence of only one trigonometric function $tan \alpha$ in the given expression urges us to find its actual value immediately,

$4+ 3\ tan \alpha=0$,

Or, $tan \alpha = -\displaystyle\frac{4}{3}$,

From * trigonometric basic function derivation principle*, we know, if we have the value of any one of the basic functions, we can derive the value of any other basic function, provided we are sure of the signs.

If the angle is acute, that is, between $0^0$ and $90^0$, all functions will be positive. But as soon as the angle, say $\alpha$ exceeds $90^0$ and moves into the range $90^0 \lt \alpha \lt 180^0$, $sin \alpha$ remains positive but $cos \alpha$ goes negative.

We always recall the signs of the trigonometric functions when the angle changes its range, by using the $sin-cos$ curve,

Instead of exactly remembering the individual function signs and also the specific important values, we derive the information using this crucial pattern of variation of $sin$ and $cos$.

This is use of * less facts more procedures approach* which reduces memory load considerably.

With the given range of $\angle \alpha$ being $90^0 \lt \angle \alpha \lt 180^0$, we determine the sign of $sin \alpha$ to be positive and the sign of the $cos \alpha$ to be negative.

**Solution 3 - Problem solving execution**

We have,

$tan \alpha=-\displaystyle\frac{4}{3}$,

Or, $1+tan^2 \alpha=sec^2 \alpha=1+\displaystyle\frac{16}{9}=\frac{25}{9}$,

Or, $sec \alpha=-\displaystyle\frac{5}{3}$,

Or, $cos \alpha=-\displaystyle\frac{3}{5}$,

Or, $sin \alpha=\sqrt{1-\displaystyle\frac{9}{25}}=\displaystyle\frac{4}{5}$.

So the given expression is,

$E=2cot \alpha - 5cos \alpha + sin \alpha$

$=-\displaystyle\frac{3}{2} +3 +\displaystyle\frac{4}{5}$

$=\displaystyle\frac{23}{10}$.

**Answer:** a: $\displaystyle\frac{23}{10}$.

**Key concepts and techniques used:** * Trigonometric function sign change with angle variation* --

**trigonometric basic function derivation principle -- sin cos curve-- less facts more procedures approach -- simplification.**

#### Trigonometric basic function derivation 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 $\ sin \theta + \ sin^2 \theta=1$, then which of the following is true?

- $\ cos \theta +\ cos^2 \theta=1$
- $\ cos^2 \theta +\ cos^3 \theta=1$
- $\ cos^2 \theta +\ cos^4 \theta=1$
- $\ cos \theta -\ cos^2 \theta=1$

**Solution 4 - Problem analysis and solving**

The nature of the given expression immediately urges us to transform it to,

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

Or, $sin\theta =1- sin^2\theta=cos^2\theta$.

As all the options consist of only $cos$ functions, as a next step we felt the need to convert the $sin\theta$ to $cos\theta$,

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

Or, $sin^2\theta=1-cos^2\theta=cos^4\theta$,

Or, $\ cos^2\theta + cos^4\theta=1$.

By target driven approach and end state analysis we find that we have reached the solution.

**Answer:** c: $cos^2\theta + cos^4\theta=1$

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

*--*

**target driven information use**

**trigonometric basic function derivation.****Problem 5.**

If $a=\sec \theta+\tan \theta$, then $\displaystyle\frac{a^2-1}{a^2+1}$ is,

- $\sec \theta$
- $\cos \theta$
- $\tan \theta$
- $\sin \theta$

**Solution 5 - Problem analysis**

As one of the important rich concepts, we recognize three pairs of trigonometric functions as * friendly function pairs*. Because of the specially close relationship between the two functions in each pair, whenever such a pair of functions appear in additive or subtractive form individually or in squares, we get a simple result. Let us explain this important well known concept.

#### Concept of friendly trigonometric function pair

Let us explain this with the first example pair of functions,

$sec\theta$ and $tan\theta$.

We have,

$sec\theta + tan\theta = \displaystyle\frac{(sec\theta +tan\theta)(sec\theta - tan\theta)}{sec\theta - tan\theta}$

$\hspace{25mm}=\displaystyle\frac{1}{sec\theta -tan\theta}$, because $sec^2\theta - tan^2\theta=1$.

The result is somewhat similar to surd rationalization.

In the same way, in the case of the second friendly function pair of $cosec\theta$ and $cot\theta$, we get,

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

The inherent friendship mechanism of the third friendly function pair, $sin\theta$ and $cos\theta$ is well known and is used very frequently.

#### Solution 5 - Problem solving execution

Recognizing the presence of a friendly pair in the given expression, we express it as,

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

Multiplying the two equations,

$a^2=\displaystyle\frac{sec\theta + tan\theta}{sec\theta -tan\theta}$,

Applying the componendo dividendo technique on the equation (subtracting 1 from both sides, again adding 1 to both sides of the original equation and taking the ratio of the two),

$\displaystyle\frac{a^2-1}{a^2+1}=\frac{tan\theta}{sec\theta}=sin\theta$.

We have reached the solution in only a few steps by applying rich trigonometry * concept of friendly trigonometric function pair* as well as

**rich algebraic technique of componendo dividendo.**This is an example of achieving what we call * elegant solution* through the process of

**efficient problem solving.****Answer:** d: $sin \theta$.

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

**concept of friendly trigonometric function pair****-- target driven approach in forming first $a^2$ and then the desired fraction -- rich algebra techniques -- componendo dividendo technique.**

**Problem 6.**

The value of $\displaystyle\frac{cot \theta + cosec \theta - 1}{cot \theta -cosec \theta +1}$ is,

- $cosec \theta - cot \theta$
- $cosec \theta + cot \theta$
- $sec \theta + cot \theta$
- $cosec \theta + tan \theta$

**Solution 6 - **Problem analysis

Examining this problem here also we detect the presence of a * friendly function pair*, $cosec \theta + cot \theta$. Only, we have to use this key pattern intelligently.

#### Solution 6 - Problem solving execution

Applying the inverse relationship of,

$cosec \theta + cot \theta =\displaystyle\frac{1}{cosec \theta - cot \theta}$, we transform the target expression to,

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

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

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

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

$=cosec \theta + cot \theta$.

Again detecting the very useful pattern of friendly trigonometric function pair and using the pattern judiciously we reach the solution elegantly in a few simple steps.

**Answer:** b: $cosec \theta + cot \theta$.

**Key concepts and techniques used:** * Key patten identification*--

*--*

**concept of friendly trigonometric function pair****-- basic algebraic concepts -- efficient simplification.**

*rich trigonometry concepts***Problem 7.**

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

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

**Solution 7 - Problem analysis and solving **

Applying componendo dividendo on the given equation we get,

$\displaystyle\frac{sin \theta}{cos \theta}=\frac{3+1}{3-1}=2$,

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

Or, $tan \theta=2$,

Or, $1+ tan^2 \theta=sec^2\theta=5$,

Or, $cos^2 \theta = \displaystyle\frac{1}{5}$.

The target expression is,

$E=sin^4 \theta - cos^4 \theta=(sin^2 \theta + cos^2 \theta)(sin^2\theta - cos^2 \theta)$

$=3cos^2 \theta=\displaystyle\frac{3}{5}$.

**Answer:** d: 1.

**Key concepts used:** Basic trigonometry concepts -- concept of friendly trigonometric function pair of $sin^2 \theta + cos^2 \theta=1$ -- * componendo dividendo technique* -- input transformation technique --

*--*

**basic trigonometric function derivation principle***.*

**multiple input use****Problem 8.**

If $asec \theta+btan \theta +c=0$, and $psec \theta +qtan \theta +r=0$, then the value of $(br-qc)^2-(pc-ar)^2$ is,

- $(aq+bp)^3$
- $(aq-bp)^3$
- $(aq+bp)^2$
- $(aq-bp)^2$

**Solution 8 - Problem analysis**

On examining the given two equations we find that there are six unknowns in addition to the two trigonometric functions. Looking at the choice values again we find the job to do is to eliminate the two trigonometric functions.

Effectively then, this is a problem in which we have to eliminate two variables from two given linear equations.

We know from our knowledge in algebra that it would always be possible to find the values of each of the two variables from two linear equations in them. Here the trigonometric functions are working as the two variables.

But how to eliminate them?

Here comes to our aid the friendly trigonometric function pair concept. If we have the values of $sec \theta$ and $tan \theta$ separately, we can easily eliminate them by the third equation which is trigonometric,

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

**Solution 8 - Problem solving execution**

Let us evaluate values of $sec \theta$ and $tan \theta$, taking $sec \theta$ first.

We have

$asec \theta+btan \theta +c=0$,

Or, $aqsec \theta + bqtan \theta + cq=0$.

Similarly multiplying the second equation by $b$,

$bpsec \theta + bqtan \theta + br=0$.

Subtracting the two,

$(aq-bp)sec \theta =(br-cq)$,

Or $sec \theta =\displaystyle\frac{br-cq}{aq-bp}$.

We can deduce similarly,

$tan \theta=\displaystyle\frac{pc-ar}{aq-bp}$.

So from $sec^2 \theta - tan^2\theta =1$,

$\displaystyle\frac{(br-cq)^2-(pc-ar)^2}{(aq-bp)^2}=1$,

Or, $(br-qc)^2-(pc-ar)^2=(aq-bp)^2$.

Though the problem looked awkward the methods followed being simple and known, solution came in a few steps.

**Answer:** d: $(aq-bp)^2$.

**Key concepts and techniques used:** End state analysis -- Deductive reasoning -- * solving of linear equations* -- basic trigonometric concepts -- basic algebraic concepts -- principle of friendly trigonometric function pairs.

**Problem 9.**

If $\alpha + \beta + \gamma=\pi$, then the value of $(sin^2 \alpha + sin^2 \beta - sin^2 \gamma)$ is,

- $2sin \alpha{sin \beta}cos \gamma$
- $2sin \alpha$
- $2sin \alpha{cos \beta}sin \gamma$
- $2sin \alpha{sin \beta}sin \gamma$

**Solution** 9 - Problem analysis

The expression $\alpha + \beta + \gamma=\pi$ and the presence of only squares of $sines$ in the target expression clearly urges ust to transform the angle expression to,

$\alpha + \beta=\pi-\gamma$ in the first step.

And then taking the $sine$ of both sides, applying compound angle concepts, squaring and simplifying we should reach the solution easily.

**Solution** 9 - Problem solving execution

$\alpha + \beta=\pi-\gamma$,

Or, $sin (\alpha + \beta)=sin (\pi-\gamma)=sin \gamma$,

Or, $(sin\alpha cos\beta + cos\alpha sin\beta)^2=sin^2 \gamma$,

Or, $sin^2\alpha cos^2\beta +2sin\alpha cos\beta cos\alpha sin\beta + cos^2\alpha sin^2\beta$

$\hspace{25mm}=sin^2\gamma$,

Or, $sin^2\alpha(1-sin^2 \beta)+sin^2\beta(1-sin^2\alpha)- sin^2 \gamma$

$\hspace{25mm}=-2sin\alpha sin\beta cos\alpha cos\beta$,

Or, $sin^2\alpha + sin^2\beta -sin^2\gamma$

$\hspace{25mm}=2sin^2\alpha sin^2\beta-2sin\alpha sin\beta cos\alpha cos\beta$,

Or, $sin^2\alpha + sin^2\beta -sin^2\gamma$

$\hspace{25mm}=-2sin\alpha sin\beta(cos\alpha cos\beta-sin\alpha sin\beta)$,

Or, $E=-2sin\alpha sin\beta cos(\alpha+\beta)$, where $E$ is the target expression,

Or, $E=-2sin\alpha sin\beta cos(\pi-\gamma)$,

Or, $E=2sin\alpha sin\beta cos \gamma$, as $cos(\pi-\theta)=-cos\theta$.

**Answer:** a: $2sin\alpha sin\beta cos \gamma$.

**Key concepts and techniques used:** Trigonometric compound angle function expressions -- friendly trigonometric function pair concept -- efficient simplification -- trigonometric function behavior with angle change.

**Note:** For detailed explanation of * compound angle function expressions* you may refer to

**Basic and rich trigonometry concepts part 2, proof of compound angle functions.****Problem 10.**

If $sin\alpha sin\beta-cos\alpha cos\beta + 1=0$, then the value of $cot\alpha tan\beta$ is,

- $-1$
- $1$
- $0$
- None of these

#### Solution 10 - Problem analysis and solving

Detecting the compound angle pattern in the target expression we transform it,

$sin\alpha sin\beta-cos\alpha cos\beta + 1=0$,

Or, $cos(\alpha+\beta)=1$,

Or, $\alpha+\beta=0^0$.

So,

$sin(\alpha+\beta)=sin\alpha cos\beta + cos\alpha sin\beta=0$,

Or, $sin\alpha cos\beta = -cos\alpha sin\beta$,

Or, $cot\alpha tan\beta=-1$.

**Answer:** a: $-1$.

**Key concepts and techniques used:** *Compound angle trigonometric functions -- basic trigonometry concepts -- end state analysis -- pattern identification.*

**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

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

#### Efficient problem solving in Trigonometry

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

**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 concepual steps, Trigonometry 8 **

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**How to solve a difficult SSC CGL level problem in a few quick steps, Trigonometry 6**

**How to solve a School Math problem in a few direct steps, Trigonometry 5**

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**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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