SSC CGL Tier II level Solution Set 11, Trigonometry 2 | SureSolv

SSC CGL Tier II level Solution Set 11, Trigonometry 2

11th SSC CGL Tier II level Solution Set, topic Trigonometry 2

SSC CGL tier 2 solution set 11 trigonometry 2

This is the 11th solution set for the 10 practice problem exercise for SSC CGL exam and 2nd on topic Trigonometry. You may refer to the 11th SSC CGL Tier II level question set and 2nd 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:

  1. Before start, you may refer to our tutorial Basic and rich Trigonometric concepts and applications or any short but good material to refresh your concepts if you so require.
  2. Answer the questions in an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes.
  3. When the time limit of 12 minutes is over, mark up to which you have answered, but go on to complete the set.
  4. At the end, refer to the answers given at the end to mark your score at 12 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.
  5. Identify and analyze the problems that you couldn't do to learn how to solve those problems.
  6. 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.
  7. 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.
  8. Give a gap before you take a 10 problem practice test again.

Important: both practice tests and mock tests must be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance.

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 section on SSC CGL to access all the valuable student resources that we have created specifically for SSC CGL, but generally for any hard MCQ test.

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You may watch the video solutions in the two-part video below.

Part 1: Q1 to Q5

Part 2: Q6 to Q10


11th solution set- 10 problems for SSC CGL Tier II exam: 2nd on Trigonometry - testing time 12 mins

Problem 1.

If $5 cos \theta +12 sin \theta=13$, and $0^0 \lt \theta \lt 90^0$, then the value of $\sin \theta$,

  1. $-\displaystyle\frac{12}{13}$
  2. $\displaystyle\frac{12}{13}$
  3. $\displaystyle\frac{5}{13}$
  4. $\displaystyle\frac{6}{13}$

Solution 1 - Problem analysis

The problem asks for value of $sin \theta$, one of the two variables from a single equation. That would have been mathermatically impossible but for the relation,

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

Now, to use this relation we have to square up the given equation. This is what we call mathematical reasoning.

Solution 1 - Useful pattern identification

At this point we give a close look at the coefficients of the three terms of the given equation and identify the distinctive relation between them,

$5^2 + 12^2=13^2$.

When we encounter such a pattern often solution is reached using the pattern. Again, to use this relation we will need to square up the given equation.

Solution 1 - Problem solving execution

Given equation,

$5 cos \theta +12 sin \theta=13$,

Squaring,

$25cos^2 \theta +120cos \theta sin \theta + 144 sin^2 \theta = 169$,

Now we notice, if we expand 169 to $169(cos^2 \theta + sin^2 \theta)$ and take all terms on RHS, we get,

$144cos^2 \theta -120cos \theta sin \theta + 25 sin^2 \theta = 0$,

Or, $(12cos \theta - 5sin \theta)^2=0$,

Or, $cot \theta = \displaystyle\frac{5}{12}$.

This is how the useful pattern of square values of 5, 12 and 13 fully exploited.

We know the by trigonometric assurance or Trigonometric basic function derivation principle that,

Once we get the value of any basic trigonometric function we can easily get the value of any other.


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

$cot \theta = \displaystyle\frac{1}{tan \theta}$.


Squaring $cot \theta$ and expanding,

$cosec^2 \theta - 1 = \displaystyle\frac{25}{144}$.

Or, $cosec^2 \theta = \displaystyle\frac{169}{144}$,

Or, $cosec \theta = \displaystyle\frac{13}{12}$, the value is positive as $\theta$ is an acute angle.

So,

$sin \theta = \displaystyle\frac{12}{13}$.

Answer: b: $\displaystyle\frac{12}{13}$.

Key concepts and techniques used: Useful pattern identification and exploitation -- basic trigonometry concepts -- trigonometric assurance -- trigonometric basic function derivation principle.

Most of the solution steps could be done in mind exploiting the pattern of $5^2 + 12^2=13^2$ in the coefficients. 

Problem 2.

The value of $(cosec \theta -sin \theta)(sec \theta - cos \theta)(tan \theta +cot \theta)$ is,

  1. 1
  2. 2
  3. 4
  4. 6

Solution 2 - Problem analysis

Knowing the power of the three most useful trigonometric relations,

$cosec^2 \theta = cot^2 \theta + 1$,

$sec^2 \theta = tan^2 \theta +1$, and

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

we look for ways to use these relations and discover that the first two component factors can be transformed to first two of these relations.

We leave the third factor as it is, as we could see how this third relation will also be transformed after we take up the immediate tasks of converting and simplifying the first two factors.

The first two relations between $cosec \theta$, $cot \theta$ and $sec \theta$, $tan \theta$ form two of the most effective trigonometric function relations in solving easy to hard MCQ based Trigonometry problems. We have named these as Friendly tigonometric function pairs.

Apart from their basic use, these friendly function pairs form more useful relationships as below.


Rich Concept of friendly trigonometric function pairs

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,

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


Once you identify a positive approach, you execute it immediately without thinking about other parts of the problem.

Solution 2 - Problem solving execution

The given target expression,

$E=(cosec \theta -sin \theta)(sec \theta - cos \theta)(tan \theta +cot \theta)$.

Inverting $sin \theta$ and $cos \theta$ respectively and simplifying,

$E=(cosec^2 \theta - 1)(sec^2 \theta - 1)(sin \theta.cos\theta)(tan \theta+cot \theta)$,

$=(cot^2 \theta)(tan^2 \theta)(sin^2 \theta + cos^2 \theta)$

$=1$.

Answer: a: 1.

Key concepts and techniques used: Useful pattern identification and exploitation -- basic trigonometry concepts -- friendly trigonometric function pairs concept -- input transformation technique -- efficient simplification.

We have identified the possibility of transforming the first two factors into friendly function pair form and accordingly transformed the input factors. Following efficienct simplification principles, we took up the first two factor transformation first, waiting for its result to come. Finally we used the byproduct of the transformations, $sin \theta.cos \theta$ in transforming the third factor to the simplest form.

All steps could easily be done in mind.

Problem 3.

If $tan A = n tan B$ and $sin A = m sin B$, then the value of $cos^2 A$ is,

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

Solution 3 - Problem analysis and solving

There are essentially four trigonometric functions, $sin A$, $cos A$ and $sin B$, $cos B$ in the given two equations. We would eliminate the other three to get the value of $cos^2 A$.

First we'll eliminate $sin A$ and $sin B$ from the first given equation by substitution of the second given equation in the first to get,

$mcos B=ncos A$.

Square it up to get,

$m^2cos^2 B=n^2cos^2 A$.

Now square up the second given equation,

$sin^2 A=m^2 sin^2 B$.

Convert the sine functions to cosines,

$1-cos^2 A=m^2(1-cos^2 B)$

$=m^2-m^2cos^2 B=m^2-n^2 cos^2 A$, from the last result.

Now it is just a step more. Rearrange the terms to get the value of $cos^2 A$ as,

$cos^2 A=\displaystyle\frac{m^2-1}{n^2-1}$

This is algebraic variable elimination method by suitable substitution, and should be the quickest.

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

Key concepts and techniques used: Variable elimination technique -- End state analysis approach -- Substitution -- Algebraic techniques.

Problem 4.

If $\theta$ is a positive acute angle and $3(sec^2 \theta + tan^2 \theta)=5$, then the value of $cos 2\theta$ is,

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

Solution 4 - Problem analysis and solving

We identify that the LHS can be reduced to a single trigonometric function variable and hence we will be able to get the value of $\theta$ directly,

$3(sec^2 \theta + tan^2 \theta)=5$

Or, $3(2sec^2 \theta -1)=5$,

Or, $2sec^2 \theta - 1 = \displaystyle\frac{5}{3}$,

Or, $cos^2 \theta=\displaystyle\frac{3}{4}$,

Or, $cos \theta = \displaystyle\frac{\sqrt{3}}{2}$, as $\theta$ is a positive acute angle,

So, $\theta = 30^0$, and

$cos 2\theta=\displaystyle\frac{1}{2}$.

Answer: c: $\displaystyle\frac{1}{2}$.

Key concepts and techniques used: End state analysis -- target driven information use -- trigonometric basic function derivation -- Frinedly trigonometric function pairs concept.

Problem 5.

If $tan \alpha = 2$, then the value of $\displaystyle\frac{cosec^2 \alpha - sec^2 \alpha}{cosec^2 \alpha+sec^2 \alpha}$ is,

  1. $-\displaystyle\frac{3}{5}$
  2. $-\displaystyle\frac{15}{9}$
  3. $\displaystyle\frac{17}{5}$
  4. $\displaystyle\frac{3}{5}$

Solution 5 - Problem solving execution by pattern identification and trigonometric function factoring

Just take out $cosec^2 \alpha$ as a factor from both numerator and denominator to convert the target expression at one stroke to,

$E=\displaystyle\frac{1-tan^2 \alpha}{1+tan^2 \alpha}$, as $\displaystyle\frac{sec^2 \alpha}{cosec^2 \alpha}=tan^2 \alpha$,

$=\displaystyle\frac{1-4}{1+4}$

$=-\displaystyle\frac{3}{5}$

If you are used to trigonometric function factoring and resultant conversion of, in this case, $sec^2 \alpha$ divided by $cosec^2 \alpha$ resulting in $tan^2 \alpha$, then this is a ten second problem.

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

We have followed the golden rule of target expression simplification first before substitution of the given value and then trigonometric function factoring.

Problem 6.

If $\sin (\theta + 30^0)=\displaystyle\frac{3}{\sqrt{12}}$ the value of $cos^2 \theta$ is,

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

Solution 6 - Problem analysis

Before going for compound angle trigonometry we would always search for a way to find out the value of the unknown angle $\theta$ directly from the given equation.

Solution 6 - Problem solving execution

$\sin (\theta + 30^0)=\displaystyle\frac{3}{\sqrt{12}}$

Or, $\sin (\theta + 30^0)=\displaystyle\frac{3}{2\sqrt{3}}$

Or, $\sin (\theta + 30^0)=\displaystyle\frac{\sqrt{3}}{2}$

So,

$\theta +30^0=60^0$, and

$\theta=30^0$, and

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

Answer: c: $\displaystyle\frac{3}{4}$.

Key concepts and techniques used: Basic trigonometry copncepts -- Trigonometric function transformation.

Problem 7.

$(1 + sec 20^0 + cot 70^0)(1 - cosec 20^0 + tan 70^0)$ is equal to,

  1. $1$
  2. $0$
  3. $-1$
  4. $2$

Solution 7 - Problem analysis 

This is a problem where we immediately detect the pattern of the relationship between the two given angles, $20^0 + 70^0=90^0$. 

Knowing the complementary relationship between $sin$ and $cos$ functions, we decide to fix $70^0$ as the target angle to which we will convert all functions,

$sec 20^0=\displaystyle\frac{1}{cos 20^0}=\frac{1}{sin 70^0}=cosec 70^0$, and

$cosec 20^0=\displaystyle\frac{1}{sin 20^0}=\frac{1}{cos 70^0}=sec 70^0$.

We have reduced number of angles from 2 to 1. This is application of two principles: Base equalization technique and Variable reduction technique - at every opportunity we try to reduce number of variables. Base angles have been equalized in addition here.

And we have used Complementary trigonometric functions concepts,

$sin (90^0 - \theta) = cos \theta$,

$cos (90^0 - \theta) = sin \theta$.

Given expression,

$E=(1 + sec 20^0 + cot 70^0)(1 - cosec 20^0 + tan 70^0)$

$=(1 + cosec 70^0 + cot 70^0)(1 - sec 70^0 + tan 70^0)$

Still simplifying,

$E=\displaystyle\frac{(sin 20^0+ cos 20^0 +1)(sin 20^0+cos 20^0 - 1)}{sin20^0cos 20^0}$

$=\displaystyle\frac{(sin 20^0 + cos 20^0)^2 - 1}{sin 20^0cos 20^0}$, basic algebraic relation, $(a+b)(a-b)=a^2-b^2$ has been used here,

$=2$.

Answer: d: 2.

Key concepts used: Pattern identification technique -- Complementary trigonometric function concepts -- Base equalization technique -- Basic trigonometry concepts --  variable reduction technique -- basic algebra concepts -- Efficient simplification.

Driving force to the solution was the need to reduce the number of angles from 2 to 1. By complementary trigonometric relation we could do that easily. Rest was efficient simplification.

Problem 8.

If $tan \theta - cot \theta =0$, and $\theta$ is a positive acute angle, then the value of $\displaystyle\frac{tan (\theta+15^0)}{tan(\theta-15^0)}$ is,

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

Solution 8 - Problem analysis

This is another problem where we would go for finding the value of $\theta$ directly from the given equation rather than think of compound angle computation.

Solution 8 - Problem solving execution

Given expression,

$tan \theta - cot \theta =0$,

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

Or, $tan^2 \theta=1$,

Or, $tan \theta=1$, as $\theta$ is positive acute angle,

So, 

$\theta=45^0$, and

$\displaystyle\frac{tan (\theta+15^0)}{tan(\theta-15^0)}$

$=\displaystyle\frac{tan (60^0)}{tan(30^0)}$

$=\displaystyle\frac{tan (60^0)}{cot(60^0)}$

$=tan^2 60^0$

$=3$.

Answer: a: $3$.

Key concepts and techniques used: Deductive reasoning -- basic trigonometry concepts.

Problem 9.

If $sec \theta - tan \theta=\displaystyle\frac{1}{\sqrt{3}}$, then the value of $sec \theta.tan \theta$ is,

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

Solution 9 - Problem analysis

This is a largely algebraic simplification. At the final stage we will subtract, $(sec \theta + tan \theta)^2 - (sec \theta - tan \theta)^2$ to get $4sec \theta.tan \theta$. This is application of End state analysis approach.

As we know how to get the value of $sec \theta+tan \theta$ by applying the rich concept of friendly trigonometric function pairs, solution path was fully visible.

Solution 9 - Problem solving execution

We know from our rich concepts,

$sec \theta + tan \theta = \displaystyle\frac{1}{sec \theta - tan \theta}=\sqrt{3}$

So,

$4sec \theta.tan \theta=3 - \displaystyle\frac{1}{3}=\displaystyle\frac{8}{3}$,

Or, $sec \theta.tan \theta=\displaystyle\frac{2}{3}$.

Answer: c: $\displaystyle\frac{2}{3}$.

Key concepts and techniques used: End state analysis approach -- friendly trigonometric function pairs rich concept -- efficient simplification.

All in mind and in quick time.


Rich Concept of friendly trigonometric function pairs

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,

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


Problem 10.

If $tan (5x - 10^0)=cot (5y+20^0)$, then the value of $x+y$ is,

  1. $15^0$
  2. $16^0$
  3. $20^0$
  4. $24^0$

Solution 10 - Problem solving execution

Just like $sin$ and $cos$ are complementary functions, $tan$ and $cot$ also are similarly a pair of complementary functions,

$cot \theta = tan (90^0 - \theta)$.

The given equation,

$tan (5x - 10^0)=cot (5y+20^0)=tan (90^0 - 5y - 20^0)$

$=tan (70^0 - 5y)$.

So,

$5x - 10^0=70^0 - 5y$

Or, $5(x+y)=80^0$,

Or, $x+y=16^0$.

Answer: b: $16^0$.

Key concepts and techniques used: Complementary trigonometric functions -- basic trigonometry concepts.

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 quick solutions of Trigonometric problems. 

You may refer to the other related resources as listed below.


Resources on Trigonometry and related topics

You may refer to our useful resources on Trigonometry and other related topics especially algebra.

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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

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