## Nineteenth SSC CGL level Solution Set, topic Trigonometry 3

This is the nineteenth solution set corresponding to the 10 practice problem exercise for SSC CGL exam and 3rd on topic Trigonometry.

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If you have not taken the corresponding test yet, first take the test by referring to * SSC CGL level Question set 19 on Trigonometry* in prescribed time and then only return to these solutions for gaining maximum benefits.

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**Part 1: Q1 to Q5**

**Part 2: Q6 to Q10**

### Nineteenth solution set- 10 problems for SSC CGL exam: 3rd on Trigonometry - testing time 12 mins

**Problem 1.**

The value of $tan1^0tan2^0tan3^0.....tan89^0$ is,

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

**Solution - Problem analysis**

From * Complementary Trigonometric functions concepts* we know,

$sin\left(\displaystyle\frac{\pi}{2} - \theta\right) = cos\theta$

$cos\left(\displaystyle\frac{\pi}{2} - \theta\right) = sin\theta$, and so,

$tan\left(\displaystyle\frac{\pi}{2} - \theta\right) = cot\theta$, where $\theta$ is acute.

With this knowledge, we find in the product a matching pair of $tan89^0$ for $tan1^0$. As $tan89^0 = cot1^0$, the product of the pair results into a 1.

Likewise, we discover the pairs, $tand2^0$ and $tan88^0$, $tan3^0$ and $tan87^0$, continuing up to the pairing of $tan44^0$ and $tan46^0$. Each of these pairs contributes 1 to the result evaluation.

Finally, only $tan45^0=1$ is left alone.

**Answer:** c: 1.

**Key concepts and techniques used:** * Identifying the useful pairing patterns* in the series of product terms -- use of

*for, $sin\theta$, $cos\theta$ and consequently the target $tan\theta$, $cot\theta$.*

**Complementary Trigonometric functions concepts****Note:** In this case the analysis and simplifying actions went hand in hand wholly in mind producing the final result in no time. Only the first identification of the patterns of result bearing pairs of $tan$ functions was needed.

**Problem 2.**

The value of $cot18^0\left(cot72^0cos^222^0 + \displaystyle\frac{1}{tan72^0sec^268^0}\right)$ is,

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

**Solution - Problem analysis**

In this second problem also using the * Complementary Trigonometric functions concepts,* we observe presence of two distinct values of angles as, $cot72^0 = tan18^0$, $tan72^0=cot18^0$ and $sec68^0=cosec22^0$. Thus only two distinct angle values of $18^0$ and $22^0$ are present. So we decide to transform all functions to these two values.

#### Solution - Simplifying actions

$E = cot18^0\left(tan18^0cos^222^0 + \displaystyle\frac{1}{cot18^0cosec^222^0}\right)$

$\hspace{5mm}=cos^222^0 + sin^222^0=1$, taking the $cot18^0$ factor inside the brackets.

**Answer:** c: 1

**Key concepts and techniques used:** Use of * Complementary Trigonometric functions concepts* -- identifying only two distinct angle values -- transformation of all terms to these two values -- simplification.

**Problem 3.**

If $asin\theta + bcos\theta =c$, then the value of $acos\theta - bsin\theta$ is,

- $\pm \sqrt{-a^2 + b^2 + c^2}$
- $\pm \sqrt{a^2 - b^2 + c^2}$
- $\pm \sqrt{a^2 - b^2 - c^2}$
- $\pm \sqrt{a^2 + b^2 - c^2}$

**Solution - Problem analysis**

The $\pm$ in the choice values indicates the solution to be reached through first squaring and then taking the square root. This is * free resource use principle* in action.

Furthermore, we detect the pattern, on squaring both the given and the target expressions, the absolute value of middle term $2absin\theta{cos\theta}$ remains same. This is **key pattern identification.**

Though theoretically there is no limit to **how many different scenarios of key patterns** can occur in solving a general problem, **this is a standard case in trigonometry** and we identify it as a **rich trigonometric concept** to be used **for significant simplification** whenever opportunity arises. We classify it under **middle term coincidence concepts **as elaborated below.

**Middle term coincidence concepts**

$(asin\theta \pm b\cos\theta)^2 = a^2sin^2\theta \pm 2absin\theta{cos\theta} + b^2cos^2\theta$, and

$(bsin\theta \pm a\cos\theta)^2 = b^2sin^2\theta \pm 2absin\theta{cos\theta} + a^2cos^2\theta$.

Though the coefficients of $sin\theta$ and $cos\theta$ in the two expressions are different, the absolute values of the middle terms coincide to the same value.

On top of it if we apply the conversion of $sin^2\theta$ to $cos^2\theta$ and vice versa, effectively **we can transpose the coefficients of $sin\theta$ and $cos\theta$.**

**Solution - Simplifying actions**

This urges us to start the solution process by squaring the given expression,

$asin\theta + bcos\theta =c$

Or, $a^2sin^2\theta + 2absin\theta{cos\theta} + b^2cos^2\theta =c^2$

Or, $a^2(1- cos^2\theta) + 2absin\theta{cos\theta} + b^2(1 - sin^2\theta) =c^2$

Or, $a^2 + b^2 - c^2 = a^2cos^2\theta - 2absin\theta{cos\theta} + b^2sin^2\theta$

Or, $(acos\theta - bsin\theta)^2 = a^2 + b^2 - c^2$

Or, $acos\theta - bsin\theta = \pm\sqrt{a^2 + b^2 - c^2}$.

**Answer:** d: $\pm \sqrt{a^2 + b^2 - c^2}$.

**Key concepts and techniques used:** Analyzing the given expression, target expression and the nature of choice values gave us the strategy to be adopted -- use of * Middle term coincidence concepts* -- the sameness of the middle term in squares of the two expressions confirmed our decision -- rest is routine algebraic simplification using the basic trigonometric relation $sin^2\theta + cos^2\theta = 1$.

**Problem 4.**

The value of $\left(\displaystyle\frac{cos^2\theta(sin\theta + cos\theta)}{cosec^2\theta(sin\theta - cos\theta)} + \displaystyle\frac{sin^2\theta(sin\theta - cos\theta)}{sec^2\theta(sin\theta + cos\theta)}\right)(sec^2\theta - cosec^2\theta) $ is,

- 1
- 2
- 3
- 4

**Solution - Problem analysis**

A general rule in Trigonometric expression simplification is, the more complex the expression looks the easier it will be to simplify it, **if you can choose the simpliflying component well.**

In this expession, as a general rule * the smaller expression should be the critical component in the simplification process* when it is applied on the larger expression.

We state this important concept in general simplication processs as,

In any simplication when two expressions are involved so that any of the two can simplify the other, always start simplification by using the smaller expression on the more complex larger expression. If required suitably transform the smaller expression first.Smaller expression precedenceconcept:

In our problem also the smaller expression needs to be transformed in $sin^2\theta$, $cos^2\theta$ form as the larger expression has similar components.

#### Solution - Simplifying actions

Transforming the smaller expression we have,

$(sec^2\theta - cosec^2\theta)=\displaystyle\frac{(sin^2\theta - cos^2\theta)}{sin^2\theta{cos^2\theta}} $

When we take this factor inside the brackets multiplying with the two complex terms, significant simplification is achieved immediately,

$\left(\displaystyle\frac{cos^2\theta(sin\theta + cos\theta)}{cosec^2\theta(sin\theta - cos\theta)} + \displaystyle\frac{sin^2\theta(sin\theta - cos\theta)}{sec^2\theta(sin\theta + cos\theta)}\right){\displaystyle\frac{(sin^2\theta - cos^2\theta)}{sin^2\theta{cos^2\theta}}} $

$=(sin\theta + cos\theta)^2 + (sin\theta - cos\theta)^2$, all other factors having been canceled out

$=2(sin^2\theta + cos^2\theta)$, the middle terms having been canceled out

$=2$

**Answer:** b: 2.

**Key concepts and techniques used:** *Identifiying key component to be used for simplifying* -- * the smaller factor is chosen as a rule* -- use of

*-- identifying the need to transform the smaller expression -- transformation and multiplication inside the brackets immediately produced the simple result -- use of basic simplifying Trigonometric and algebraic relations $a^2 - b^2 = (a + b)(a - b)$ and $sin^2\theta + cos^2\theta = 1$.*

**smaller expression precedence concept****Problem 5.**

$\displaystyle\frac{tan\theta}{1 - cot\theta} + \displaystyle\frac{cot\theta}{1 - tan\theta}$ is equal to,

- $1 - tan\theta -cot\theta$
- $1 + tan\theta + cot\theta$
- $1 - tan\theta + cot\theta$
- $1 + tan\theta - cot\theta$

**Solution - Problem analysis**

We identify only one unique variable in the expression as, $tan\theta =\displaystyle\frac{1}{cot\theta}$.

$tan\theta=x$ is taken as the unique variable and all the terms are converted into this unique variable using * Reverse substitution technique* of rich algebraic concepts. Also by this process we reduce the number of variables from two to 1, a use of

*which is a rich algebraic technique.*

**Variable reduction technique,****Solution - **Simplifying actions

Thus the target expression,

$E = \displaystyle\frac{x}{1 - \displaystyle\frac{1}{x}} + \displaystyle\frac{\displaystyle\frac{1}{x}}{1 - x}$

$\hspace{5mm}=\displaystyle\frac{x^2}{x - 1} - \displaystyle\frac{1}{x(x - 1)}$

$\hspace{5mm}=\displaystyle\frac{x^3 - 1}{x(x - 1)}$

Now we will use the algebraic expression, $x^3 - 1 = (x - 1)(x^2 + x + 1)$. Thus,

$E = \displaystyle\frac{x^2 + x + 1}{x}$

$\hspace{5mm}=x + 1 + \displaystyle\frac{1}{x}$

$\hspace{5mm}=1 + tan\theta + cot\theta$

**Answer:** b: $1 + tan\theta + cot\theta$.

**Key concepts and techniques used:** * Reducing number of variables to 1, *use of

*using*

**Variable reduction technique --***transforming the target expression to an algebraic expression -- combining the two terms immediately produced the important simplifying expression $x^3 - 1$ in the numerator -- simplification and resubstitution of trigonometric variable $tan\theta$.*

**reverse substitution technique****Note:** The advantage of using $x=tan\theta$ is, *ease of dealing with known algebraic forms for simplification.* **The trigonometric forms may hide the algebraic form by cluttering**. Here we have used another important algebraic rich concept of * Variable reduction technique*. This is a powerful technique that must be applied by analyzing complex expressions.

**Problem 6.**

If $tan\theta = \displaystyle\frac{sin\alpha - cos\alpha}{sin\alpha + cos\alpha}$ then $sin\alpha + cos\alpha$ is,

- $\pm \sqrt{2} sin\theta$
- $\pm \sqrt{2} cos\theta$
- $\pm \displaystyle\frac{1}{\sqrt{2}} cos\theta$
- $\pm \displaystyle\frac{1}{\sqrt{2}} sin\theta$

**Solution - **Problem analysis

As always, we do not resort to deductions straightaway, but analyze the problem, quickly identifying its special form and strategy to be adopted.

With this problem we notice the **presence of the target expression in the denominator in the RHS of the given expression. **This is use of * End state analysis approach* of comparing the target and the given expression to identify useful similarities. Thus we decide to go

**for eliminating the numerator**$(sin\alpha - cos\alpha)$

**keeping the expression in the denominator intact.**

The **second decision** that we take is to **use squaring and then taking the square root as is evident from the form of the choice values. **This is * free resource use principle* in action.

With these decisions we start our simplification actions.

#### Solution - Simplifying actions

Squaring both sides of the equation and adding 1,

$tan^2\theta + 1 = \displaystyle\frac{(sin\alpha - cos\alpha)^2}{(sin\alpha + cos\alpha)^2} + 1$

$sec^2\theta = \displaystyle\frac{(sin\alpha - cos\alpha)^2 + (sin\alpha + cos\alpha)^2}{(sin\alpha + cos\alpha)^2}$.

The numerator in the RHS is in the form of a standard rich concept which we name as * Middle term cancellation concept* as detailed below.

#### Rich algebraic concept of Middle term cancellation

When squares of sum and subtraction expressions in two variables are added, the middle terms are canceled out leaving twice the sum of squares of the two variables.

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

When applied to trigonometry we have,

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

$=2(sin^2\theta + cos^2\theta)$, the middle terms $2sin\theta{cos\theta}$ cancel out

$=2$, a much more simplified result.

Usually when we use algebraic simplification techniques in trigonometry, greater simplification results are achieved.

Thus from the last result we have,

$sec^2\theta = \displaystyle\frac{2}{(sin\alpha + cos\alpha)^2}$,

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

Or, $sin\alpha + cos\alpha = \pm\sqrt{2}cos\theta$.

**Answer:** b: $\pm \sqrt{2} cos\theta$.

**Key concepts and techniques used:** Identifying the presence of target expression in the given expression by the use of * End state analysis approach* -- adoping the strategy of preserving it -- again identifying from the choice values the need for squaring and taking square root by

*-- lastly after squaring the two terms in the equation and adding 1 to both sides to get the highly simplifying trigonometric expression $(sin\alpha - cos\alpha)^2 + (sin\alpha + cos\alpha)^2$ that conforms to the*

**free resource use principle***and simplifies to just the value of 2 -- thus the expression $(sin\alpha - cos\alpha)$ in the numerator is eliminated leaving only the target expression in the denominator -- final simplification.*

**middle term cancellation concept****Problem 7.**

If $cos^2\alpha - sin^2\alpha = tan^2\beta$, then $cos^2\beta - sin^2\beta = $

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

**Solution - problem analysis**

Noticing $tan^2\beta$ on the RHS of the given expression, we decide to use its basic relation $tan^2\beta = sec^2\beta - 1$ to get $cos^2\beta$ and then from it the second term in the target expression $sin^2\beta$.

**Solution - simplifying actions**

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

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

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

$sin^2\beta = 1 - cos^2\beta = 1 - \displaystyle\frac{1}{2cos^2\alpha} = \displaystyle\frac{2cos^2\alpha - 1}{2cos^2\alpha}$.

So the target expression,

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

$\hspace{10mm}=\displaystyle\frac{2(1 - cos^2\alpha)}{2cos^2\alpha}$

$\hspace{10mm}=tan^2\alpha$

**Answer:** a: $tan^2\alpha$.

**Key concepts and techniques used:** One of the most used and unfailing concept that we frequently apply in Trigonometry is * Trigonomteric function transformation principle*,

If we have one of the basic trigonometric functions such as $sin\theta$, $cos\theta$, or $tan\theta$, specially in square form, we can always get any other basic function from the basic relations between the squares of the basic functions, such as $sin^2\theta + cos^2\theta = 1$ and, $tan^2\theta = sec^2\theta -1$.

This principle in Trigonometry is so basic and is so frequently used, that we named it simply as, **Trigonomteric function transformation principle.**

Using this concept first deriving the first term of the target expression $cos^2\beta$, simplifying the RHS on the way -- next getting the second term $sin^2\beta$ from the $cos^2\beta$ itself -- final simplification.

**Problem 8.**

If $tan\alpha=ntan\beta$, and $sin\alpha = msin\beta$ then $cos^2\alpha$ is,

- $\displaystyle\frac{m^2 - 1}{n^2 - 1}$
- $\displaystyle\frac{m^2 + 1}{n^2 + 1}$
- $\displaystyle\frac{m^2}{n^2 + 1}$
- $\displaystyle\frac{m^2}{n^2}$

**Solution - problem analysis**

From the **target expression and the choice value**s it becomes evident that we need to eliminate $\beta$ functions and we can do that by finding the same $\beta$ function from the two given expressions in terms of purely $\alpha$ functions and equating the two.

In the **second stage of analysis,** we needed to **decide on the $\beta$ function that we would equate.**

In this case, observing that the second expression is in terms of $sin$ function, we know we can convert it to $cot$ function by inverting, squaring and subtracting 1 from it.

$cosec^2\beta - 1 = cot^2\beta$.

Looking at the first given function now, choice of $cot\beta$ as the target function becomes a firm one, as we can convert the $tan$ functions in the first expression without adding any extra term in the process.

This last statement is extremely important in any simplification process. Let us restate it in tems of a rich algebraic concept * Minimum addition of extra terms*,

In any algebraic simplification step, choose the step that adds minimum number of extra terms to the expressions.

This can be stated as a subset of more general rich algebraic technique, * Term reduction technique*,

In any algebraic simplification, always take the path that will reduce the number of terms to its minimum at that point. This will invariably be the shortest path to the solution.

Usually this powerful concept will lead you to the final result quickest.

**Solution - simplifying actions**

As we have decided $cot\beta$ to be the target function we directly take up the first expression to get $cot^2\beta$ in terms of purely $\alpha$ functions.

$tan\alpha=ntan\beta$,

Or, $cot\beta = \displaystyle\frac{n}{tan\alpha}$

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

We proceed to convert the $sec^2\alpha$ to $cos^2\alpha$ with an eye fixed on the target expression where we have only $cos^2\alpha$.

Taking similar actions on the second expression we have,

$sin\alpha = msin\beta$,

Or, $cosec^2\beta = \displaystyle\frac{m^2}{sin^2\alpha} = \displaystyle\frac{m^2}{1 - cos^2\alpha}$,

In the same vein we convert the $sin^2\alpha$ to $cos^2\alpha$ as that is in the target expression.

Again converting $cosec^2\beta$ to $cot^2\beta$,

$cot^2\beta + 1 = \displaystyle\frac{m^2}{1 - cos^2\alpha}$,

Or, $cot^2\beta = \displaystyle\frac{m^2}{1 - cos^2\alpha} - 1 = \displaystyle\frac{m^2 - 1 + cos^2\alpha}{1 - cos^2\alpha}$

Equating the two as planned, we get,

$\displaystyle\frac{n^2cos^2\alpha}{1 - cos^2\alpha} = \displaystyle\frac{m^2 - 1 + cos^2\alpha}{1 - cos^2\alpha}$,

Or, $n^2cos^2\alpha = m^2 - 1 + cos^2\alpha$

Or, $cos^2\alpha = \displaystyle\frac{m^2 - 1}{n^2 - 1}$,

**Answer:** a: $\displaystyle\frac{m^2 - 1}{n^2 - 1}$.

**Key concepts and techniques used:** By analysis of the target expression and the choice values identifying the need to eliminate $\beta$ functions -- using the * algebraic concept of minimum addition of extra terms* and analyzing the two given expressions,

**taking the decision to transform both the given expressions in terms of $cot^2\beta$**and

**equating the two**-- simplifying actions according to the decisions taken.

**Problem 9.**

If $A$, $B$ and $C$ are the three angles of a triangle, then the incorrect relation among the following is,

- $cos\displaystyle\frac{A + B}{2} = sin\displaystyle\frac{C}{2}$
- $sin\displaystyle\frac{A + B}{2} = cos\displaystyle\frac{C}{2}$
- $cot\displaystyle\frac{A + B}{2} = tan\displaystyle\frac{C}{2}$
- $tan\displaystyle\frac{A + B}{2} = sec\displaystyle\frac{C}{2}$

**Solution** - Problem analysis

In a triangle, the sum of three angles is $180^0$, that is,

$A + B + C = 180^0$,

Or, $\displaystyle\frac{A + B}{2} = \left(\displaystyle\frac{\pi}{2} - \displaystyle\frac{C}{2}\right)$.

The same will be the relations for the other pairs of angles $(B + C)$ and $(C + A)$.

With these relations, applying the * complementary trigonometric functions concept* of

$sin\left(\displaystyle\frac{\pi}{2} - \theta\right) = cos\theta$

$cos\left(\displaystyle\frac{\pi}{2} - \theta\right) = sin\theta$

$tan\left(\displaystyle\frac{\pi}{2} - \theta\right) = cot\theta$, and

$cot\left(\displaystyle\frac{\pi}{2} - \theta\right) = tan\theta$,

we would easily be able to identify the odd wrong relation out of the four.

#### Solution - Simplifying actions

Just observing the choices we find that the first three relations follow the * complementary trigonometric functions concept *perfectly. But in the fourth relation, $sec$ is not the complementary function of $tan$, rather $cot$ is. So this is the wrong one.

**Answer:** d: $tan\displaystyle\frac{A + B}{2} = sec\displaystyle\frac{C}{2}$.

**Key concepts and techniques used:** Analysis of the problem transforms it for the application of * complementary trigonometric functions concept *and consequent easy identification of the wrong relation.

**Problem 10.**

If $\theta$ is a positive acute angle and $tan2\theta{tan3\theta} = 1$ then the value of $\left(2cos^2\displaystyle\frac{5\theta}{2} - 1\right)$ is,

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

#### Solution - Problem analysis

From the target expression the complexity of the angle involved made us fairly certain that we have to find the actual value of $\theta$ from the given expression and then substitute the value in the target expression.

With this conjecture in mind when we examine the given expression we can easily visualize the possibility for forming a $tan-cot$ relation and get the value of $\theta$ by applying the frequently used **complementary trigonometric functions concept.**

#### Solution - Simplifying actions

$tan2\theta{tan3\theta} = 1$,

Or, $tan2\theta=cot3\theta=tan(90^0 - 3\theta)$

Or, $2\theta = 90^0 - 3\theta$,

Or, $5\theta = 90^0$,

Or, $\theta = 18^0$, as expected.

Substituting this value in target expression we get,

$\left(2cos^2\displaystyle\frac{5\theta}{2} - 1\right) = 2cos^245^0 - 1 = 0$.

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

**Key concepts and techniques used:** From complexity of angle form in target expression deciding to get the actual value of $\theta$ from the given expression -- with this conjecture in mind it was easy to tranform the given expression into a $tan-cot$ relation and then on to a $tan-tan$ relation applying **complementary trigonometric functions concept.**

**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.

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