SSC CGL level Solution Set 56, Trigonometry 5

Trigonometry aptitude questions SSC CGL set 56: Quick solutions

Trigonometry aptitude questions with solutions SSC CGL Set 56. 10 selected questions solved in simple steps by basic and advanced concepts and techniques.

If not taken already, take the test at 56th SSC CGL question set and 5th on Trigonometry before you go through the solution.

---

Trigonometry aptitude questions with solutions SSC CGL Set 56 - 10 selected questions - testing time was 12 mins

Problem 1.

The maximum value of $(2\sin \theta + 3\cos\theta)$ is,

1. $1$
2. $2$
3. $\sqrt{13}$
4. $\sqrt{15}$

Solution - Problem analysis and execution

To find the maximum value, the given expression is converted to an expression involving a single trigonometric function using the compound angle relation of,

$c\sin (\theta +\alpha)=c\left(\sin {\theta}\cos \alpha + cos {\theta}\sin \alpha\right)$.

The coefficients $a$ and $b$ are substituted with,

$a=c\cos \alpha$, $b=c\sin \alpha$.

The value of the new coefficient $c$, is then,

$c=\sqrt{a^2+b^2}$.

It is in fact the hypotenuse of a right triangle in which $a$ and $b$ are the other two sides and $\angle \alpha$ is the base angle.

As maximum value of $\sin (\theta +\alpha)$ is 1, the maximum value of the given expression turns out to be,

$c=\sqrt{a^2+b^2}$.

In this case, $a=2$ and $b=3$, and so the desired maximum value is,

$\sqrt{2^2+3^2}=\sqrt{13}$.

Answer: c: $\sqrt{13}$.

Key concepts and techniques used: Conversion to single term function -- maxima minima of trigonometric expressions -- substitution technique -- rich trigonometry concepts.

Note: For detailed treatment on maxima minima of trigonometric expressions, refer to the tutorial session on Trigonometry concepts part 3, maxima (or minima) of trigonometric expressions.

Problem 2.

Find the minimum value of $9\tan^2 \theta + 4\cot^2 \theta$,

1. $15$
2. $6$
3. $9$
4. $12$

Solution - Problem analysis

As $\tan$ or $\cot$ functions both have undefined maximum value, the given expression can have only a minimum value.

How to decide which method of maxima minima finding technique to be followed?

In this case the two trigonometric functions when multiplied with each other cancel out leaving a purely numeric result as these two form a pair of inverse trigonometric functions as well as their powers are same. This is the guiding criteria for applying the well known AM GM inequality technique.

The AM (Arithmetic Mean) GM (Geometric Mean) inequality technique simply says,

For a set of values,

$\text{AM} \geq \text{GM}$.

For a two term expression, $9\tan^2 \theta + 4\cot^2 \theta$, the AM of the two terms taken individually as two values, is,

$\text{AM} = \displaystyle\frac{9\tan^2 \theta + 4\cot^2 \theta}{2}$,

and the GM of the two terms is,

$\text{GM}=\sqrt{\left(9\tan^2 \theta)(4\cot^2 \theta\right)}=6$.

So applying the inequality condition,

$\displaystyle\frac{9\tan^2 \theta + 4\cot^2 \theta}{2} \geq 6$,

Or, $9\tan^2 \theta + 4\cot^2 \theta \geq 12$.

The given expression can only then be greater than or equal to 12. So its minimum value is 12.

Answer: d: $12$.

Key concepts and techniques used: Problem analysis -- Problem solving strategy -- Maxima minima of trigonometric expressions -- AM GM inequality -- inverse trigonometric functions.

Problem 3.

If $\text{cosec} \theta -\cot \theta= \displaystyle\frac{7}{2}$, the value of $\text{cosec} \theta$ is,

1. $\displaystyle\frac{49}{28}$
2. $\displaystyle\frac{53}{28}$
3. $\displaystyle\frac{47}{28}$
4. $\displaystyle\frac{51}{28}$

Solution - Problem analysis and solving

From friendly function pair concept we know,

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

So from the given value, $\text{cosec} \theta -\cot \theta= \displaystyle\frac{7}{2}$, we have,

$\text{cosec} \theta +\cot \theta= \displaystyle\frac{2}{7}$.

Adding the two equations,

$2\text{cosec} \theta = \displaystyle\frac{2}{7}+\displaystyle\frac{7}{2}$

$=\displaystyle\frac{53}{14}$,

Or, $\text{cosec} \theta = \displaystyle\frac{53}{28}$.

Answer: b: $\displaystyle\frac{53}{28}$.

Key concepts and techniques used: Friendly function pairs concept -- Basic algebra concepts.

Problem 4.

If $\tan^2 \alpha=1 +2\tan^2 \beta$, where $\alpha$ and $\beta$ both are positive acute angles, find the value of $\sqrt{2}\cos \alpha - \cos \beta$.

1. $0$
2. $\sqrt{2}$
3. $-1$
4. $1$

Solution - Problem analysis and solving

The single given equation must be modified to produce the result. How to modify it?

Examining the equation, it seems to be unbalanced. The shortage of 1 on the RHS as well as on the LHS became visible in no time, because whenever there are two $\tan^2$s in the input and two $\cos$'s in the output, the relation of $sec^2$ with $\tan^2$ comes to mind.

So we add 1 to both sides, getting,

$1+\tan^2 \alpha=2 +2\tan^2 \beta$,

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

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

So,

$\sqrt{2}\cos \alpha = \cos \beta$,

Or, $\sqrt{2}\cos \alpha - \cos \beta=0$.

Answer: a: $0$.

Key concepts and techniques used: Expression balance assessment -- Missing element introduction technique - End state analysis approach.

Problem 5.

The value of $152(\sin 30^0 + 2\cos^2 45^0 + 3\sin 30^0 +4\cos^2 45^0 + ....+17\sin 30^0+18\cos^2 45^0)$ is,

1. an irrational number
2. a rational number but not an integer
3. an integer but not a perfect square
4. the perfect square of an integer

Solution - Problem analysis

By a quick look at the terms we find,

$\sin 30^0=\frac{1}{2}$ in every odd term and also,

$\cos^2 45^0=\frac{1}{2}$ in every even term.

Then $\frac{1}{2}$ can be taken out of the brackets as a common factor thus dividing 152 by 2 with a result of 76 and leaving a sum of natural numbers inside the brackets,

$1+2+3+.....+17+18$.

This is nothing but the sum of first 18 natural numbers which we evaluate quickly to be $9\times{17}+18=9\times{19}$,

Mechanism: $9$ is the middle term of the first 17 natural numbers and is the average of these 17 numbers, we add another 18 to get the sum of first 18 natural numbers.

Finally then the given equation evaluates to,

$76\times{9}\times{19}=36\times{19}\times{19}$, a perfect square of an integer, $6\times{19}=114$.

Answer: d: the perfect square of an integer.

Key concepts and techniques used: Common pattern identification -- Problem transformation to finding sum of first 18 natural numbers -- Efficient simplification, we didn't carry out the multiplication but kept the factors visible for later use for finding the nature of the number -- basic factors and multiples concept-- basic trigonometry concepts.

Problem 6.

If $\tan \theta + \cot \theta=2$, then the value of $\tan^n \theta + \cot^n \theta$ ($0^0 \lt \theta \lt 90^0$, and $n$ an integer) is,

1. $2$
2. $2^{n+1}$
3. $2^n$
4. $2n$

Solution - Problem analysis and solving

As the functions are directly raised to $n$th power without any extra help, each of the functions must be 1. Let us verify our hypothesis.

The given equation,

$\tan \theta + \cot \theta=2$,

Or, $\tan \theta + \displaystyle\frac{1}{\tan \theta}=2$,

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

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

Or, $\tan \theta = \cot \theta =1$.

So,

$\tan^n \theta + \cot^n \theta=2$.

Answer: a: $2$.

Key concepts and techniques used: Deductive reasoning -- Input transformation -- basic trigonometric concepts.

Problem 7.

If $\displaystyle\frac{\sin \theta}{1+\cos \theta} + \displaystyle\frac{\sin \theta}{1-\cos \theta} = 4$, the value of $\cot \theta + \sec \theta$ is,

1. $\sqrt{3}$
2. $\displaystyle\frac{\sqrt{3}}{7}$
3. $\displaystyle\frac{\sqrt{3}}{5}$
4. $\displaystyle\frac{5}{\sqrt{3}}$

Solution - problem analysis and solving

We visualize the possibility of eliminating the denominators of each term separately and then form a simple addition,

$\displaystyle\frac{\sin \theta}{1+\cos \theta} + \displaystyle\frac{\sin \theta}{1-\cos \theta} = 4$,

Or, $\displaystyle\frac{1}{\text{cosec} \theta+\cot \theta} + \displaystyle\frac{1}{\text{cosec} \theta-\cot \theta}=4$, we bring down the $\sin \theta$ from the numerator to the denominator by dividing it

Or, $(\text{cosec} \theta - \cot \theta) + (\text{cosec} \theta + \cot \theta)=4$, using friendly function pair concept,

Or, $2\text{cosec} \theta = 4$,

Or, $\sin \theta = \frac{1}{2}$, that is, $\theta=30^0$.

So,

$\cot \theta = \cot 30^0=\sqrt{3}$, and,

$\sec \theta = \sec 30^0=\displaystyle\frac{2}{\sqrt{3}}$.

Thus we have the target expression,

$\cot \theta + \sec \theta=\sqrt{3}+\displaystyle\frac{2}{\sqrt{3}}=\displaystyle\frac{5}{\sqrt{3}}$

Answer: d: $\displaystyle\frac{5}{\sqrt{3}}$.

Key concepts used: Key pattern identification -- Denominator elimination -- Friendly trigonometric function pair -- Basic trigonometry concepts.

Problem 8.

If $\sin \theta + \cos \theta =\sqrt{2}$ with $\angle \theta$ a positive acute angle, then the value of $\tan \theta + \sec \theta$ is,

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

Solution - problem analysis and solving

The given equation is,

$\sin \theta + \cos \theta =\sqrt{2}$

Squaring both sides,

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

Or, $2\sin {\theta}\cos \theta = 1$,

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

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

Or, $\sin \theta - \cos \theta = 0$,

Or, $\sin \theta=\cos \theta=\displaystyle\frac{1}{\sqrt{2}}$, from $\sin \theta + \cos \theta = \sqrt{2}$.

Thus,

$\tan \theta=1$, and,

$\sec \theta=\sqrt{2}$.

So,

$\tan \theta + \sec \theta = \sqrt{2}+1$.

Multiplying the numerator and denominator of RHS by $(\sqrt{2}-1)$,

$\tan \theta +\sec \theta =\displaystyle\frac{1}{\sqrt{2}-1}$.

Answer: b: $\displaystyle\frac{1}{\sqrt{2}-1}$.

Key concepts and techniques used: Deductive reasoning -- basic trigonometric concepts -- surd rationalization -- basic algebra concepts.

Note; Without any deduction also, from the given expression, value of $\theta=45^0$ can be guesssed by testing the given expression.

Problem 9.

If $p=a\sec {\theta}\cos \alpha$, $q =b\sec {\theta}\sin \alpha$, and $r =c\tan \theta$, then the value of $\displaystyle\frac{p^2}{a^2} +\displaystyle\frac{q^2}{b^2}-\displaystyle\frac{r^2}{c^2}$ is,

1. 0
2. 1
3. 4
4. 5

Solution - problem analysis and solving

Analyzing the target expression with respect to the given expression we decide to form the three individual terms of the target expression from the three input expressions by input transformation.

The first given equation is,

$p=a\sec {\theta}\cos \alpha$,

Or, $\displaystyle\frac{p^2}{a^2}=\sec^2 {\theta}\cos^2 \alpha$.

Similarly from the second and third given expressions we get,

$\displaystyle\frac{q^2}{b^2}=\sec^2 {\theta}\sin^2 \alpha$, and

$\displaystyle\frac{r^2}{c^2}=\tan^2 {\theta}$.

Adding the first two transformed expressions,

$\displaystyle\frac{p^2}{a^2}+ \displaystyle\frac{q^2}{b^2}=\sec^2 {\theta}\cos^2 \alpha +\sec^2 {\theta}\sin^2 \alpha$,

Or, $\displaystyle\frac{p^2}{a^2}+ \displaystyle\frac{q^2}{b^2}=\sec^2 \theta$

$=1+\tan^2 \theta$

$=1+\displaystyle\frac{r^2}{c^2}$

So,

$\displaystyle\frac{p^2}{a^2} +\displaystyle\frac{q^2}{b^2}-\displaystyle\frac{r^2}{c^2}=1$

Answer: b: 1.

Key concepts and techniques used: End state analysis comparing target with givens comparing similarities -- input transformation -- basic trigonometric concepts -- basic algebra concepts.

Problem 10.

$\displaystyle\frac{\sin^2 \theta}{\cos^2 \theta}+\displaystyle\frac{\cos^2 \theta}{\sin^2 \theta}$ is equal to,

1. $\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta}$
2. $\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta} -2$
3. $\displaystyle\frac{1}{\tan^2 \theta - \cot^2 \theta}$
4. $\displaystyle\frac{\sin^2 \theta}{\cot \theta - \sec \theta}$

Solution - Problem analysis

As combining the two terms results in $\sin^4 \theta+\cos^4 \theta$ as numerator, this option seems to lead to a more complex solution path and so is rejected.

On brief examination of the choice values, the $-2$ term in the Option 2 attracts attention and prompts us more towards adding and subtracting 2 to the expression, and breaking down the problem into individual term simplification.

The 2 added is allocated 1 to each term to eliminate the numerator and simplify the numerator considerably.

Instead of Denominator elimination technique here we eliminate the numerator to the value of 1 by adding 1 to each term.

Solution - Problem solving execution

Adding and subtracting 2 from the expression and allocating 1 to each term for addition we obtain,

$\displaystyle\frac{\sin^2 \theta}{\cos^2 \theta}+\displaystyle\frac{\cos^2 \theta}{\sin^2 \theta}$

$=\left[\displaystyle\frac{\sin^2 \theta}{\cos^2 \theta}+1\right] + \left[\displaystyle\frac{\cos^2 \theta}{\sin^2 \theta}+1\right] -2$

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

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

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

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

Answer: b: $\displaystyle\frac{1}{\sin^2 {\theta}\cos^2 \theta}-2$.

Key concepts and techniques used: Problem analysis, combining the the two terms seems to lead to a more complex path, and so is rejected -- End state analysis, as the term $-2 $ in choice 2 prompted us to add 2 and subtract 2 -- Problem breakdown technique, individual terms are simplified first -- Individual term simplification, often simplification of each term in an expression speeds up the solution considerably -- Introduction of new simplifying terms, the two 1s added to the two terms to simplify each considerably -- Numerator elimination, the numerator of each term is converted to 1, effectively eliminating it -- Basic trigonometry concepts -- Basic algebra concepts -- Rich algebra concepts and techniques.

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.

---

Guided help on Suresolv Trigonometry

To get the best results out of the extensive range of articles on Quantitative Aptitude Trigonometry problem solving, follow the guide,

Trigonometry Reading and Practice Guide for SSC CGL, SSC CGL Tier II and Other Competitive exams.

Basically, it is how to read and practice Suresolv Trigonometry guide.

The guide list of articles will be always UPTODATE.

---