## Second SSC CGL level Solution Set, topic Trigonometry

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

It is emphasized here that answering in MCQ test is not at all the same as answering in a school test where you need to derive the solution in perfectly elaborated steps.

In MCQ test instead, you need basically to deduce the answer in shortest possible time and select the right choice. None will ask you about what steps you followed.

Based on our analysis and experience we have seen that, for accurate and quick answering, the student

- must have complete understanding of the basic concepts of the topics
- is adequately fast in mental math calculation
- should try to solve each problem using the most basic concepts in the specific topic area and
- does most of the deductive reasoning and calculation in his head rather than on paper.

Actual problem solving happens in item 3 and 4 above. But how to do that?

You need to use your **your problem solving abilities** only. There is no other recourse.

If you have not taken the corresponding test yet, first take the test by referring to * SSC CGL level Question Set 2 on Trigonometry* in prescribed time and then only return to these solutions for gaining maximum benefits.

### Second solution set- 10 problems for SSC CGL exam: topic Trigonometry - time 20 mins

**Q1.** If $0^0 < \theta < 90^0$ and $2sin^2\theta + 3cos\theta = 3$ then the value of $\theta$ is,

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

**Solution:** You identify the given equation as a second order equation mixed in $sin\theta$ and $cos\theta$.

You reason that to find the value of $\theta$ you must know the value of $sin\theta$ or $cos\theta$, and that you can get from a second order quadratic equation in either $sin\theta$ or $cos\theta$, but not both mixed up. It implies that you must convert either $sin\theta$ to $cos\theta$ or $cos\theta$ to $sin\theta$.

This is Deductive reasoning starting from the target, that is, the end point. This type of analysis falls under the powerful End State Analysis Approach. Most of the tricky math problems can be quickly solved by intelligent use of this approach.

Thus you decide at this point to convert the $sin^2\theta$ to $cos^2\theta$ by the most important trigonometric identity, $sin^2\theta + cos^2\theta = 1$. This is the first transformation on the given equation, and we get given equation as,

\begin{align} E = & 2(sin^2\theta + cos^2\theta) \\ & -2cos^2\theta + 3cos\theta = 3 \end{align}

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

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

This is simple algebra where you have to use your knowledge in factorization of quadratic equation at the least. Hint - noticing $2cos^2\theta$ in the second order term and $1$ in the third numeric term, you straightway decide one factor to be $(2cos\theta - 1)$.

Now only you use the first condition in the problem. In case of second root, $cos\theta=1$ and $\theta=0^0$ that violates the given condition. The first root $cos\theta=\frac{1}{2}$ gives $\theta=60^0$, the answer.

If you find it difficult to remember the values of $sin\theta$ and $cos\theta$ for various values of $\theta$, you just have to remember the following picture depicting variation of $sin\theta$ and $cos\theta$ for various values of $\theta$.

From the figure it can never be forgotten that $cos\theta=1$ when $\theta=0^0$ and its value reduces to $0$ when value of $\theta$ goes on increasing to $90^0$. In between there are two significant values of $cos\theta$, namely $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ at $\theta$ values either $30^0$ or $60^0$. As $\frac{1}{2}$ is smaller than $\frac{\sqrt{3}}{2}$, and also as value of $cos\theta$ reduces from 1 to 0, it follows that $cos60^0=\frac{1}{2}$.

This is direct application of the Principle of Less facts and more procedure.

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

**Key concept used:** Elimination of $sin\theta$ -- factorization -- use of given condition. To remember values of $cos\theta$ without loading the memory use of powerful problem solving Principle of Less facts more procedures.

**Q2.** If $sin\theta=\frac{a}{\sqrt{a^2 + b^2}}$, then the value of $cot\theta$ will be,

- $\frac{b}{a}$
- $\frac{a}{b}$
- $\frac{a}{b} + 1$
- $\frac{b}{a} + 1$

**Solution:** We reason, to find the value of $cot\theta$, we need values of $sin\theta$ and $cos\theta$, one of which is given. Using the identity, $ sin^2\theta + cos^2\theta = 1$ we can mentally arrive at the result,

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

Or, $cos\theta = \frac{b}{\sqrt{a^2 + b^2}}$

So, $cot\theta = \frac{b}{a}$.

**Answer:** Option a : $\frac{b}{a}$ .

**Key concept used:** Find the unknown element in the target $cot\theta$ using given $sin\theta$.

**Alternative method:** We remember that there is a relationship between $sin\theta$ and $cot\theta$ via $cosec\theta$ using the identity, $cosec^2\theta = 1 + cot^2\theta$.

We simply inverse the given equation and square it to get,

\begin{align} cosec^2\theta & = 1 + cot^2\theta \\ & = \frac{a^2 + b^2}{a^2} \\ & = 1 + \frac{b^2}{a^2} \end{align}

which gives in turn,

$cot\theta = \frac{b}{a}$.

Can you spot the difference between the two methods? Apparently both are quick ways to reach the solution. Think for a few minutes.

The first method worked from the very basics and the second used a rich concept. Usually we prefer to solve a problem using the most basic concept. That is fastest. But in this case, with repeated use, the two rich concepts involving $tan^2\theta + 1 = sec^2\theta$ and $cot^2\theta + 1 = cosec^2\theta$ can very well be used as first level resource in establishing relationship between $sec\theta$, $tan\theta$ and $cosec\theta$, $cot\theta$.

**Q3.** If $tan\theta=\frac{3}{4}$ and $0<\theta<\frac{\pi}{2}$ and $25xsin^2\theta{cos\theta}=tan^2\theta$, then the value of $x$ is,

- $\frac{7}{64}$
- $\frac{9}{64}$
- $\frac{3}{64}$
- $\frac{5}{64}$

**Solution:** In any problem of this type, you examine first the target expression. Here target expression is the third one as it is the most complex. The second expression here is a condition and the first is a simple identity. Best plan is to use these simpler given information to simplify the most complex expression.

This again is another important principle of problem solving. If two facts or expressions are given, one much more complex than the other, always know that, you will start simplifying the second using the first. This principle we call as Principle of relative complexity. We use the less complex information as a tool to simplify the more complex information.

In the second expression, we cancel out the $sin^2\theta$ on both sides and take the $cos\theta$ on the RHS to get,

$25x = sec\theta{sec^2\theta}$

Now we use the simpler expression $tan\theta=\frac{3}{4}$ to derive,

$sec^2\theta=1 + tan^2\theta=\frac{25}{16}$.

Substituting we get,

$25x=\frac{25}{16}sec\theta$,

Or, $x = \frac{sec\theta}{16}$

Now is the time to use the second condition in the expression, $sec\theta = \pm{\frac{5}{4}=\frac{5}{4}}$.

Finally we then get $x=\frac{5}{64}$.

**Answer:** Option d: $\frac{5}{64}$.

**Key concept used:** Simplifying more complex trigonometric expression -- using the given simpler information suitably.

**Q4.** If $xsin\theta - ycos\theta = \sqrt{x^2 + y^2}$ and $\frac{cos^2\theta}{a^2} + \frac{sin^2\theta}{b^2} = \frac{1}{x^2 + y^2}$ then,

- $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
- $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$
- $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
- $\frac{x^2}{b^2} - \frac{y^2}{a^2} = 1$

**Solution:** At first the expressions seem to be a little complex, but on examining the first expression (as usual the resource to be used as a tool) against the second expression and the choices, you will find first, there is no square root.

So without hesitation you undertake to square the expression and simplify,

$x^2sin^2\theta + y^2cos^2\theta$

$\qquad - 2xysin\theta{cos\theta} = x^2 + y^2$

Transposing and collecting $sin^2\theta$ and $cos^2\theta$ it further simplifies to,

$x^2cos^2\theta + y^2sin^2\theta + 2xysin\theta{cos\theta} = 0$

Or, $(xcos\theta + ysin\theta)^2 = 0$

Or, $xcos\theta = -ysin\theta$

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

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

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

Or, $cos^2\theta = 1 - sin^2\theta = \frac{y^2}{x^2 + y^2}$

Now your target is to eliminate $cos\theta$ and $sin\theta$, by substituting values of $cos^2\theta=\frac{y^2}{x^2 + y^2}$, and $sin^2\theta=\frac{x^2}{x^2 + y^2}$ in the second expression, giving,

$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.

**Answer:** Option b: $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.

**Key concept used:** Identifying no square root -- squaring and simplifying -- eliminating $sin\theta$ and $cos\theta$ in second expression.

**Q5.** The value of $sin^21^0 + sin^23^0 + sin^25^0 + ...$

$... + sin^287^0 + sin^289^0$ is,

- $22$
- $22\frac{1}{2}$
- $23$
- $22\frac{1}{4}$

**Solution:** We must have the basic knowledge that,

$sin\theta = cos(90^0 - \theta)$ for positive $\theta$ less than or equal to $90^0$. Using this knowledge we can form pairs sum of each of which will be 1 as below,

\begin{align} sin^21^0 + sin^289^0 & = cos^289^0 + sin^289^0 \\ & = 1 \end{align}

Question now is how many terms does the expression have? Without using any formula, just test for a similar expression ending with $sin^27^0$ and another with the next one $sin^29^0$. We know the first to have 4 terms and the second 5 terms. Knowing this we try the conjecture that number of terms for the first = $\frac{7 + 1}{2} = 4$ and for the second, $\frac{9 + 1}{2} = 5$ both tallying with our direct knowledge. Thus confirming our conjecture, we use the often used Principle of induction to get the number of terms in the given expression = $\frac{89 + 1}{2} = 45$.

With 45 terms, we get 22 pairs giving sum as 22 and a single term in the middle, the 23rd term. The second term has angle $2\times{(2 - 1)} + 1 = 3^0$, the third, $2\times{(3 - 1)} + 1 = 5^0$. It tallies. Using induction again, we get the angle of the middle term = $2\times{(23 - 1)} + 1 = 45^0$.

As $sin45^0 = \frac{1}{\sqrt{2}}$, we get the given sum = $22\frac{1}{2}$.

**Answer:** Option b: $22\frac{1}{2}$.

**Key concept used:** Identifying the useful pattern of pairing using the relationship $sin\theta = cos(90^0 - \theta)$ and of course $sin^2\theta + cos^2\theta = 1$ --- finding number of terms and the middle term.

**Q6.** The minimum value of $cos^2\theta + sec^2\theta$ is,

- 0
- 1
- 2
- 3

** Solution:** The value of $cos\theta$ ranges from -1 to 1 intermediate values being less than 1. That is $-1\leq\cos\theta\leq1$. This gives, $0\leq\cos^2\theta\leq1$.

Thus assuming $cos^2\theta = \frac{1}{x}$, we get $x\geq1$, that is, a positive real number with minimum value 1. Substituting in the given expression, we get,

$E=\frac{1}{x} + x = \frac{x^2 + 1}{x}$, where $x\geq1$.

As $x\geq1$ and numerator grows much faster than the denominator with increasing value of $x$, the expression will have minimum value when $x$ has minimum value of 1. So minimum value of expression will be 2.

**Answer:** Option c : 2.

**Key concept used:** Transforming the expression in such a way that positive valued variables greater than 1 form the numerator and the denominator allowing comparison of the growth of the two with increasing value of $x$ -- using the knowledge of how changing variable values affect an expression.

**Q7.** If $cos\theta + sec\theta = 2$ $(0^0\leq{\theta}\leq{90^0})$ then the value of $cos{10}\theta + sec{11}\theta$ is,

- 0
- 1
- 2
- -1

** Solution:** After initial examination we find the target expression a little hard to break. Recalling that simpler initial expression is to be used for evaluating more complex expression, we concentrate on the first expression.

$cos\theta + sec\theta = 2$

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

Or, $cos\theta=1$, giving $\theta=0^0$. So given expression = 2.

**Answer:** Option c: 2.

**Key concept used:** Simplification of given expression automatically resolves the problem of $10\theta$ and $11\theta$.

**Q8.** If $tan\theta=\frac{3}{4}$ and $\theta$ is acute then, $cosec\theta$ is equal to,

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

**Solution:** $cot\theta=\frac{4}{3}$, or, $cot^2\theta = cosec^2\theta - 1 = \frac{16}{9}$, or, $cosec^2\theta = \frac{25}{9}$, or, $cosec\theta = \frac{5}{3}$.

**Answer:** Option a: $\frac{5}{3}$.

**Key concept used:** Recognition and use of $cosec^2\theta$ and $cot^2\theta$ relation.

**Q9.** If $\displaystyle\frac{sin\theta + cos\theta}{sin\theta - cos\theta} = 3$ then the numerical value of $sin^4\theta - cos^4\theta$ is,

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

**Solution: **Adding 1 to both sides, subtracting 1 from both sides and taking ratio of the two we get, $tan\theta = 2$.

Target expression,

\begin{align} E & = (sin^2\theta + cos^2\theta)(sin^2\theta - cos^2\theta) \\ & = (sin^2\theta - cos^2\theta) \\ & = 1 - 2cos^2\theta \end{align}

Again $tan\theta = 2$, or, $tan^2\theta = 4 = sec^2\theta - 1$, or, $sec^2\theta = 5$, or, $cos^2\theta = \frac{1}{5}$

Thus $E = 1 - \frac{2}{5} = \frac{3}{5}$

**Answer:** Option c: $\frac{3}{5}$.

**Key concept used:** Simplification of given expression to get value of $tan\theta$, and then $cos^2\theta$ - simplifying target expression in terms of $cos^2\theta$. Simplification using componendo dividendo technique which is a standard technique when the numerator and denominator are in this form. Getting $tan\theta$ we can always get $sec\theta$ and hence $cos\theta$.

**Q10.** The minimum value of $2sin^2\theta + 3cos^2\theta$ is,

- 0
- 3
- 2
- 1

**Solution:** Target expression,

$E = 2 + cos^2\theta$

Minimum value of $cos^2\theta$ being 0, minimum value of target expression is 2.

**Answer:** Option c: 2.

**Key concept used:** Simplification of given expression -- minimum value of $cos^2\theta$.

**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 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.

#### SSC CGL Tier II level question and solution sets on Trigonometry

**SSC CGL Tier II level Solution set 12 Trigonometry 3, questions with solutions**

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

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

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

**SSC CGL Tier II level Solution set 7 Trigonometry 1**

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

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**SSC CGL level Question Set 16 on Trigonometry**

**SSC CGL level Question Set 2 on Trigonometry**

**SSC CGL level Solution Set 2 on Trigonometry**

#### 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**