Trigonometry Questions for SSC CPO with Answers and Solution 1

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SSC CPO Solved question Set 3, Trigonometry 1

Trigonometry questions answers solutions for SSC CPO 1

1st Set of 10 Trigonometry questions for SSC CPO with answers and quick solutions

Trigonometry problem solving questions for SSC CPO Set 1. Verify from answers and learn how to solve quick from solutions. Previous year questions.

The solved question set contains,

  1. Question set on Trigonometry for SSC CPO to be answered in 15 minutes (10 chosen questions)
  2. Answers to the questions, and
  3. Quick conceptual solutions to the questions.

Take the timed test, verify correctness from answers and learn to solve quick from solutions.


Trigonometry questions for SSC CPO Set 1 - answering time 15 mins

Q1. If $\tan\left(\displaystyle\frac{\theta}{2}\right)\tan\left(\displaystyle\frac{2\theta}{5}\right)=1$, what is the value (in degrees) of $\theta$?

  1. $120^0$
  2. $100^0$
  3. $45^0$
  4. $90^0$

Q2. If $\text{cosec }\theta-\text{cot }\theta=\displaystyle\frac{7}{2}$, then the value of $\text{cosec }\theta$ is,

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

Q3. If $\sin (A-B)=\sin A\cos B-\cos A\sin B$, then $\sin 15^0$ will be,

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

Q4. If $\text{cosec}^2 \theta +\text{cot}^2 \theta=7$, what is the value (in degrees) of $\theta$?

  1. $30$
  2. $60$
  3. $15$
  4. $45$

Q5. What is the simplified value of $\left[(1+\text{sec }2\theta)\tan^2 \theta\right]+1$?

  1. $\text{cosec }\theta$
  2. $1$
  3. $\text{sec }2\theta$
  4. $\cos 2\theta$

Q6. If $\cos (\theta+31^0)=\sin 47^0$, what is the value of $\sin 5\theta$?

  1. $\displaystyle\frac{1}{2}$
  2. $\displaystyle\frac{\sqrt{3}}{2}$
  3. $\displaystyle\frac{1}{\sqrt{2}}$
  4. $0$

Q7. What is the value of $\displaystyle\frac{\text{sec }\theta}{\tan \theta+\text{cot }\theta}$

  1. $\sin \theta$
  2. $\cos \theta$
  3. $\tan \theta$
  4. $\text{cot }\theta$

Q8. If $\text{cosec}^2 \theta=\displaystyle\frac{25}{16}$ and $\theta$ is acute, then what is the value of $(\sqrt{\tan \theta+\text{cot }\theta})$?

  1. $\displaystyle\frac{\sqrt{3}}{2}$
  2. $\displaystyle\frac{12}{\sqrt{3}}$
  3. $\displaystyle\frac{2\sqrt{3}}{3}$
  4. $\displaystyle\frac{5}{2\sqrt{3}}$

Q9. What is the simplified value of $1+\text{cot }A.\text{cot }\left(\displaystyle\frac{A}{2}\right)$?

  1. $\sin^2\left(\displaystyle\frac{A}{2}\right)$
  2. $\cos \left(\displaystyle\frac{A}{2}\right)$
  3. $\displaystyle\frac{1}{2}\text{cosec}^2 \left(\displaystyle\frac{A}{2}\right)$
  4. $\cos A$

Q10. What is the simplified value of $\displaystyle\frac{2\sin^3 \theta-\sin \theta}{\cos \theta-2\cos^3 \theta}$?

  1. $\cos \theta$
  2. $\tan \theta$
  3. $\sin \theta$
  4. $\text{cot }\theta$

Answers to the Trigonometry questions for SSC CPO Set 1

Q1. Answer: Option b: $100^0$.

Q2. Answer: Option d: $\displaystyle\frac{53}{28}$.

Q3. Answer: Option d: $\displaystyle\frac{\sqrt{3}-1}{2\sqrt{2}}$.

Q4. Answer: Option a: $30$.

Q5. Answer: Option c: $\text{sec }2\theta$.

Q6. Answer: Option b: $\displaystyle\frac{\sqrt{3}}{2}$.

Q7. Answer: Option a: $\sin \theta$.

Q8. Answer: Option d: $\displaystyle\frac{5}{2\sqrt{3}}$.

Q9. Answer: Option c: $\displaystyle\frac{1}{2}\text{cosec}^2 \left(\displaystyle\frac{A}{2}\right)$.

Q10. Answer: Option b: $\tan \theta$.


Solutions to the Trigonometry questions for SSC CPO Set 1 - answering time 15 mins

Q1. If $\tan\left(\displaystyle\frac{\theta}{2}\right)\tan\left(\displaystyle\frac{2\theta}{5}\right)=1$, what is the value (in degrees) of $\theta$?

  1. $120^0$
  2. $100^0$
  3. $45^0$
  4. $90^0$

Solution 1: Immediate solution by converting one $\tan$ to $\text{cot}$, taking it to the RHS and applying complementary angle trigonometric relation between $\tan$ and $\text{cot}$

Take one factor from LHS to RHS and apply complementary angle trigonometric relation,

$\tan\left(\displaystyle\frac{\theta}{2}\right)\tan\left(\displaystyle\frac{2\theta}{5}\right)=1$,

Or, $\tan\left(\displaystyle\frac{\theta}{2}\right)=\text{cot }\left(\displaystyle\frac{2\theta}{5}\right)$

$=\tan\left(\displaystyle\frac{\pi}{2}-\displaystyle\frac{2\theta}{5}\right)$.

So,

$\displaystyle\frac{\theta}{2}=\displaystyle\frac{\pi}{2}-\displaystyle\frac{2\theta}{5}$,

Or, $\displaystyle\frac{\theta}{2}+\displaystyle\frac{2\theta}{5}=\displaystyle\frac{\pi}{2}=90^0$,

Or, $\displaystyle\frac{9\theta}{10}=90^0$,

Or, $\theta=\displaystyle\frac{10}{9}\times{90^0}=100^0$.

Answer: Option b: $100^0$.

Key concepts used: Key pattern identification -- Complementary angle trigonometric relations -- Solving in mind.

Note: This method will always work for a product of two factors of same trigonometric function but different angles equated to 1. This is the key pattern.

Q2. If $\text{cosec }\theta-\text{cot }\theta=\displaystyle\frac{7}{2}$, then the value of $\text{cosec }\theta$ is,

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

Solution 2: Immediate solution by friendly trigonometric mutually inverse function pair concept

As, $\text{cosec}^2\theta-\text{cot}^2\theta=1$,

$\text{cosec }\theta-\text{cot }\theta=\displaystyle\frac{1}{\text{cosec }\theta+\text{cot }\theta}=\frac{7}{2}$,

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

Add the two equations now to eliminate $\text{cot }\theta$,

$2\text{cosec }\theta=\displaystyle\frac{7}{2}+\displaystyle\frac{2}{7}=\displaystyle\frac{53}{14}$.

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

Answer: Option d: $\displaystyle\frac{53}{28}$.

Key concepts used: Mutually inverse friendly trigonometric function pair concept -- Simple algebraic variable elimination -- Solving in mind.

Q3. If $\sin (A-B)=\sin A\cos B-\cos A\sin B$, then $\sin 15^0$ will be,

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

Solution 3: Quick solution by key pattern identification of suitable values $A=60^0$ and $B=45^0$ so that $A-B=15^0$

Identify the key pattern that with $A=60^0$ and $B=45^0$, you will get $\sin 15^0$ on LHS and on RHS all function values known,

$\sin (60^0-45^0)=\sin 60^0\cos 45^0-\cos 60^0\sin 45^0$

Or, $\sin 15^0=\displaystyle\frac{\sqrt{3}}{2}\left(\displaystyle\frac{1}{\sqrt{2}}\right)-\displaystyle\frac{1}{2}\left(\displaystyle\frac{1}{\sqrt{2}}\right)=\displaystyle\frac{\sqrt{3}-1}{2\sqrt{2}}$.

Answer: Option d: $\displaystyle\frac{\sqrt{3}-1}{2\sqrt{2}}$.

Key concepts used: Key pattern identification of two suitable angles, difference of which is $15^0$ and values of $\sin$ and $\cos$ functions for both the angles known -- Solving in mind.

Q4. If $\text{cosec}^2 \theta +\text{cot}^2 \theta=7$, what is the value (in degrees) of $\theta$?

  1. $30$
  2. $60$
  3. $15$
  4. $45$

Solution 4: Quick solution by converting $\text{cot}^2 \theta$ to $\text{cosec}^2 \theta$, variable reduction technique

In Algebra, we have used the powerful Variable reduction technique,

In an expression if you reduce the number of variables by at least 1, solution can be reached quickly.

In this case of two variables in the LHS of the given equation, convert $\text{cot}^2 \theta$ to $\text{cosec}^2 \theta$,

$\text{cosec}^2 \theta +\text{cot}^2 \theta=7$,

Or, $\text{cosec}^2 \theta +\text{cosec}^2 \theta-1=7$,

Or, $2\text{cosec}^2 \theta=8$,

Or, $\sin^2 \theta=\displaystyle\frac{1}{4}$,

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

Answer: Option a: $30$.

Key concepts used: Key pattern identification -- Friendly trigonometric function pair relation, $\text{cosec}^2 \theta-1=\text{cot}^2 \theta$ -- Variable reduction technique -- Solving in mind.

Q5. What is the simplified value of $\left[(1+\text{sec }2\theta)\tan^2 \theta\right]+1$?

  1. $\text{cosec }\theta$
  2. $1$
  3. $\text{sec }2\theta$
  4. $\cos 2\theta$

Solution 5: Solve quickly by trigonometric multiple angle relations

We'll use the trigonometric multiple angle relation,

$\cos 2\theta=\cos^2 \theta-\sin^2 \theta$.

First identify the opportunity to simplify the target expression,

$(1+\text{sec }2\theta\tan^2 \theta)+1$

$=\text{sec }2\theta\tan^2 \theta +(1+\tan^2 \theta)$

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

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

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

The function of $\text{sec }2\theta$ has been to neutralize $\tan^2 \theta$ as well as the extra unity term.

Answer: Option c: $\text{sec }2\theta$.

Key concepts used: Trigonometric multiple angle relations -- Basic trigonometric angle relation $\sin^2 \theta+\cos^2 \theta=1$ -- Multiple submultiple angle trigonometric functions.

Q6. If $\cos (\theta+31^0)=\sin 47^0$, what is the value of $\sin 5\theta$?

  1. $\displaystyle\frac{1}{2}$
  2. $\displaystyle\frac{\sqrt{3}}{2}$
  3. $\displaystyle\frac{1}{\sqrt{2}}$
  4. $0$

Solution 6: Solving in mind by trigonometric complementary angle relation

Applying trigonometric complementary angle relation, the given equation is transformed to,

$\cos (\theta+31^0)=\sin 47^0=\cos (90^0-47^0)=\cos 43^0$.

So,

$\theta+ 31^0=43^0$,

Or, $\theta=12^0$, and,

$5\theta=60^0$.

So,

$\sin 5\theta=\sin 60^0=\displaystyle\frac{\sqrt{3}}{2}$.

Answer: Option b: $\displaystyle\frac{\sqrt{3}}{2}$.

Key concepts used: Trigonometric complementary angle relation -- Trigonometric function values for specific angles -- Solving in mind.

Q7. What is the value of $\displaystyle\frac{\text{sec }\theta}{\tan \theta+\text{cot }\theta}$

  1. $\sin \theta$
  2. $\cos \theta$
  3. $\tan \theta$
  4. $\text{cot }\theta$

Solution 7: Quick solution by converting $\text{cot }\theta$ to $\tan \theta$ and Variable reduction technique

Convert $\text{cot }\theta$ to $\tan \theta$ to reduce number of variables by 1,

$\displaystyle\frac{\text{sec }\theta}{\tan \theta+\text{cot }\theta}$

$=\displaystyle\frac{\text{sec }\theta}{\tan \theta+\displaystyle\frac{1}{\tan \theta}}$

$=\displaystyle\frac{\text{sec }\theta\tan \theta}{1+\tan^2 \theta}$

$=\displaystyle\frac{\text{sec }\theta\tan \theta}{\text{sec}^2 \theta}$

$=\displaystyle\frac{\tan \theta}{\text{sec } \theta}$

$=\sin \theta$

Answer: Option a: $\sin \theta$.

Key concepts used: Key pattern identification -- Variable reduction by converting $\text{cot }\theta$ to $\tan \theta$ -- Solving in mind.

Q8. If $\text{cosec}^2 \theta=\displaystyle\frac{25}{16}$ and $\theta$ is acute, then what is the value of $(\sqrt{\tan \theta+\text{cot }\theta})$?

  1. $\displaystyle\frac{\sqrt{3}}{2}$
  2. $\displaystyle\frac{12}{\sqrt{3}}$
  3. $\displaystyle\frac{2\sqrt{3}}{3}$
  4. $\displaystyle\frac{5}{2\sqrt{3}}$

Solution 8: Quick solution by principle of deriving any trigonometric function value from a given trigonometric function value

We will evaluate first $\text{cot }\theta$ from given value of $\text{cosec }\theta$,

$\text{cosec}^2 \theta=\displaystyle\frac{25}{16}$,

Or, $1+\text{cot}^2 \theta=\displaystyle\frac{25}{16}$,

Or, $\text{cot}^2 \theta=\displaystyle\frac{9}{16}$,

Or, $\text{cot }\theta=\displaystyle\frac{3}{4}$, as $\theta$ is acute all six trigonometric functions are positive.

So,

$\tan \theta=\displaystyle\frac{1}{\text{cot }\theta}=\frac{4}{3}$.

Add the two,

$\tan \theta+\text{cot }\theta=\displaystyle\frac{4}{3}+\displaystyle\frac{3}{4}=\displaystyle\frac{25}{12}$,

Or, $\sqrt{\tan \theta+\text{cot }\theta}=\displaystyle\frac{5}{2\sqrt{3}}$.

Answer: Option d: $\displaystyle\frac{5}{2\sqrt{3}}$.

Key concepts used: Deriving trigonometric function value of first $\text{cot }\theta$ from given value and then of $\tan \theta$ -- Trigonometric basic function derivation principle -- Solving in mind.

Q9. What is the simplified value of $1+\text{cot }A.\text{cot }\left(\displaystyle\frac{A}{2}\right)$?

  1. $\sin^2\left(\displaystyle\frac{A}{2}\right)$
  2. $\cos \left(\displaystyle\frac{A}{2}\right)$
  3. $\displaystyle\frac{1}{2}\text{cosec}^2 \left(\displaystyle\frac{A}{2}\right)$
  4. $\cos A$

Solution 9: Solve quick by trigonometric multiple submultiple angle relations

The angle $\left(\displaystyle\frac{A}{2}\right)$ is a submultiple angle of angle $A$. For ease of use, we will convert the submultiple angle problem to a multiple angle problem that is a type of compound angle problem.

To do this we will use $\theta=\left(\displaystyle\frac{A}{2}\right)$ to transform the given target equation to,

$E=1+\text{cot }A.\text{cot }\left(\displaystyle\frac{A}{2}\right)$

$=1+\text{cot }2\theta\text{cot }\theta$.

We will use the multiple angle relations,

$\sin 2\alpha=2\sin \alpha\cos \alpha$, and,

$\cos 2\alpha=\cos^2 \alpha - \sin^2 \alpha=2\cos^2 \alpha -1$.

Use these relations in the target expression,

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

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

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

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

$=\displaystyle\frac{1}{2}\text{cosec}^2 \theta$

$=\displaystyle\frac{1}{2}\text{cosec}^2\left(\displaystyle\frac{A}{2}\right)$.

Answer: Option c: $\displaystyle\frac{1}{2}\text{cosec}^2 \left(\displaystyle\frac{A}{2}\right)$.

Key concepts used: Trigonometric multiple submultiple angle relations -- Basic trigonometric relations.

Q10. What is the simplified value of $\displaystyle\frac{2\sin^3 \theta-\sin \theta}{\cos \theta-2\cos^3 \theta}$?

  1. $\cos \theta$
  2. $\tan \theta$
  3. $\sin \theta$
  4. $\text{cot }\theta$

Solution 10: Quick solution by key pattern identification of factoring out $\sin \theta$ from numerator and $\cos \theta$ from denominator

The numerator and denominator of target expression are factorized to the simplified result,

$E=\displaystyle\frac{2\sin^3 \theta-\sin \theta}{\cos \theta-2\cos^3 \theta}$

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

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

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

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

Substituting these results in the numerator and denominator,

$E=\tan \theta.\displaystyle\frac{\sin^2 \theta-\cos^2 \theta}{\sin^2 \theta-\cos^2 \theta}$

$=\tan \theta$.

Answer: Option b: $\tan \theta$.

Key concepts used: Key pattern identification - Factoring out to simplify numerator and denominator to known forms -- Basic trigonometric relation of $\sin^2 \theta+\cos^2 \theta=1$ -- Solving in mind.

End note

Observe that, each of the problems could be quickly and cleanly solved in minimum number of steps using special key patterns and methods in each case.

This is the hallmark of quick problem solving:

  • Concept based pattern and method formation, and,
  • Identification of the key pattern and use of the method associated with it. Every special pattern has its own method, and not many such patterns are there.

Important is the concept based pattern identification and use of quick problem solving method.


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