## 65th SSC CGL level Question Set, topic Trigonometry 6

This is the 65th question set for the 10 practice problem exercise for SSC CGL exam and 6th on topic Trigonometry. You will find the answers to the questions along with list of related readings including the detailed conceptual solutions after the questions.

#### Before taking the test it is recommended that you refer to the tutorials

**Tutorial on Basic and rich concepts in Trigonometry and its applications.**

**Tutorial on Basic and rich concepts in Trigonometry part 2, proof of compound angle functions**

**Tutorial on Trigonometry concepts part 3, maxima or minima of Trigonometric expressions**

**You may also refer to the related guideline:**

* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****generally for any hard MCQ test.**

If you like,you mayto get latestsubscribecontent on competitive examspublished in your mail as soon as we publish it.

Now set the stopwatch alarm and start taking this test. It is not difficult.

### 65th question set- 10 problems for SSC CGL exam: 6th on Trigonometry - testing time 12 mins

**Problem 1.**

If $2-cos^2 \theta=3sin \theta{cos \theta}$, where $sin \theta \neq cos \theta$, the value of $tan \theta$ is,

- $0$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{1}{3}$

**Problem 2.**

If $sin \theta + cos \theta =\sqrt{2}cos(90^0- \theta)$, then $cot \theta$ is,

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

**Problem 3.**

If $(a^2-b^2)sin \theta + 2abcos \theta=a^2+b^2$ then the value of $tan \theta$ is,

- $\displaystyle\frac{1}{2ab}(a^2+b^2)$
- $\displaystyle\frac{1}{2}(a^2-b^2)$
- $\displaystyle\frac{1}{2}(a^2+b^2)$
- $\displaystyle\frac{1}{2ab}(a^2-b^2)$

**Problem 4.**

The value of $sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right) - 2tan^2 \theta$ is,

- 4
- 0
- 2
- 1

**Problem 5.**

If $cot \theta + cosec \theta =3$, and $\theta$ an acute angle, the value of $cos \theta$ is,

- $1$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{3}{4}$

**Problem 6.**

If $xcos \theta - sin \theta=1$, then the value of $x^2-(1+x^2)sin \theta$ is,

- $1$
- $0$
- $2$
- $-1$

**Problem 7.**

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

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

**Problem 8.**

If $3sin \theta + 5cos \theta =5$, ($0\lt \theta \lt 90^0$), then the value of $5sin \theta-3cos \theta$ will be,

- 1
- 2
- 5
- 3

**Problem 9.**

If $tan \theta = \displaystyle\frac{1}{\sqrt{11}}$, and $0\lt \theta \lt 90^0$, then the value of $\displaystyle\frac{cosec^2 \theta - sec^2 \theta}{cosec^2 \theta + sec^2 \theta}$ is,

- $\displaystyle\frac{5}{6}$
- $\displaystyle\frac{3}{4}$
- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{6}{7}$

**Problem 10.**

If $tan^2 \theta=1-e^2$, then the value of $sec \theta + tan^3 \theta{cosec \theta}$ is equal to,

- $(2+e^2)^{\frac{3}{2}}$
- $(2+e^2)^{\frac{1}{2}}$
- $(2-e^2)^{\frac{3}{2}}$
- $(2-e^2)^{\frac{1}{2}}$

The answers to the questions are given below, but you will find the * detailed conceptual solutions* to these questions in

*.*

**SSC CGL level Solution Set 65 on Trigonometry 6**### Answers to the questions

Problem 1. **Answer:** b: $\displaystyle\frac{1}{2}$.

Problem 2. **Answer:** a: $\sqrt{2}-1$.

Problem 3. **Answer:** d: $\displaystyle\frac{1}{2ab}(a^2-b^2)$.

Problem 4. **Answer:** Option c: 2.

Problem 5. **Answer:** c: $\displaystyle\frac{4}{5}$.

Problem 6. **Answer:** a: $1$.

Problem 7. **Answer:** b: $cos \displaystyle\frac{\theta}{2}$.

Problem 8. **Answer:** d: 3.

Problem 9. **Answer:** a: $\displaystyle\frac{5}{6}$.

Problem 10. **Answer:** c: $(2-e^2)^{\frac{3}{2}}$.

### 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.

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