## Sixteenth SSC CGL level Question Set on Trigonometry

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

We found from our analysis of the Trigonometry problems that this topic is built on a * small set of basic and rich concepts.* That's why it's possible to solve any problem in this topic area fast and quick, following elegant problem solving methods if you are used to applying problem solving techniques based on related basic and rich subject concepts.

We have tried to show you how this can be done in the solution set. But please, first take this test in prescribed time.

### Sixteenth Question set- 10 problems for SSC CGL exam: topic Trigonometry - time 12 mins

**Problem 1.**

The simplified value ofÂ $(sec\theta - cos\theta)^2 + (cosec\theta - sin\theta)^2 - (cot\theta - tan\theta)^2$ is,

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

**Problem 2.**

If $\displaystyle\frac{sin\theta + cos\theta}{sin\theta - cos\theta} = \frac{5}{4}$, then the value of $\displaystyle\frac{tan^2\theta + 1}{tan^2\theta - 1}$ will be,

- $\displaystyle\frac{41}{40}$
- $\displaystyle\frac{40}{41}$
- $\displaystyle\frac{25}{16}$
- $\displaystyle\frac{41}{9}$

**Problem 3.**

If $sin\theta + cosec\theta =2$, then the value of $sin^{100}\theta + cosec^{100}\theta$ is,

- 100
- 3
- 2
- 1

**Problem 4.**

The greatest value of $sin^4\theta + cos^4\theta$ is,

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

**Problem 5.**

If $\displaystyle\frac{sin\theta}{x} = \displaystyle\frac{cos\theta}{y}$, then $sin\theta - cos\theta$ is,

- $x - y$
- $\displaystyle\frac{x - y}{\sqrt{x^2 + y^2}}$
- $\displaystyle\frac{y - x}{\sqrt{x^2 + y^2}}$
- $x + y$

**Problem 6.**

If $tan\theta - cot\theta = 0$ find the value of $sin\theta + cos\theta$.

- $\sqrt{2}$
- $0$
- $1$
- $2$

**Problem 7.**

If $sin21^0 = \displaystyle\frac{x}{y}$ then $sec21^0 - sin69^0$ is,

- $\displaystyle\frac{y^2}{x\sqrt{y^2 - x^2}}$
- $\displaystyle\frac{x^2}{y\sqrt{y^2 - x^2}}$
- $\displaystyle\frac{x^2}{y\sqrt{x^2 - y^2}}$
- $\displaystyle\frac{y^2}{x\sqrt{x^2 - y^2}}$

**Problem 8.**

If $\displaystyle\frac{sec\theta+ tan\theta}{sec\theta - tan\theta}=\displaystyle\frac{5}{3}$ then $sin\theta$ is,

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

**Problem 9.**

If $(1 + sin A)(1 + sin B)(1 + sin C) = (1 - sin A)(1 - sin B)( 1 - sin C)$, then the expression on each side of the equation equals,

- $1$
- $tan A.tan B.tan C$
- $cos A.cos B.cos C$
- $sin A.sin B.sin C$

**Problem 10.**

If $\theta = 60^0$, then, $\displaystyle\frac{1}{2}\sqrt{1 + sin\theta} + \displaystyle\frac{1}{2}\sqrt{1 - sin\theta}$ is,

- $cos\displaystyle\frac{\theta}{2}$
- $cot\displaystyle\frac{\theta}{2}$
- $sec\displaystyle\frac{\theta}{2}$
- $sin\displaystyle\frac{\theta}{2}$

You will find the detailed conceptual solutions to these questions in * SSC CGL level Solution Set 16 on Trigonometry*.

**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. But compared to difficulties of purely algebraic problem solving, trigonometry problems are simpler because by applying a few basic and rich trigonometric concepts along with algebraic concepts elegant solutions are reached faster.

Watch the **video solutions** in the **two-part video.**

**Part 1: Q1 to Q5**

**Part 2: Q6 to Q10**

### Answers to the questions

**Problem 1. Answer:** Option d: $1$.

**Problem 2. Answer:** Option a : $\displaystyle\frac{41}{40}$.

**Problem 3. Answer:** Option c: 2.

**Problem 4. Answer:** Option a: $1$.

**Problem 5. Answer:** Option b: $\displaystyle\frac{x - y}{\sqrt{x^2 + y^2}}$.

**Problem 6. Answer:** Option a : $\sqrt{2}$.

**Problem 7. Answer:** Option b: $\displaystyle\frac{x^2}{y\sqrt{y^2 - x^2}}$.

**Problem 8. Answer:** Option d: $\displaystyle\frac{1}{4}$.

**Problem 9. Answer:** Option c: $cos A.cos B.cos C$.

**Problem 10. Answer:** Option a: $cos\displaystyle\frac{\theta}{2}$.

### 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 18 Trigonometry 4, questions with solutions**

**SSC CGL Tier II level Question Set 18 Trigonometry 4, questions with answers**

**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**

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

#### SSC CGL level question and solution sets in Trigonometry

**SSC CGL level Solution set 82 on Trigonometry 8**

**SSC CGL level Question set 82 on Trigonometry 8**

**SSC CGL level solution set 77 on Trigonometry 7**

**SSC CGL level question set 77 on Trigonometry 7**

**SSC CGL level Solution Set 65 on Trigonometry 6**

**SSC CGL level Question Set 65 on Trigonometry 6**

**SSC CGL level Solution Set 56 on Trigonometry 5**

**SSC CGL level Question Set 56 on Trigonometry 5**

**SSC CGL level Solution Set 40 on Trigonometry 4**

**SSC CGL level Question Set 40 on Trigonometry 4**

**SSC CGL level Solution Set 19 on Trigonometry**

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

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

**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**