## 12th SSC CGL Tier II level Question Set, topic Trigonometry 3

This is the 12th question set for the 10 practice problem exercise for SSC CGL exam and 3rd on topic Trigonometry.

The **answers to the questions** and **link to the detailed solutions** are given at the end.

**Recommendation:** Before taking the test you should refer to the tutorials on,

* Basic and rich concepts in Trigonometry and its applications,* and,

**Basic and rich algebraic concepts for elegant solutions of SSC CGL problems.**

The test should be used as a **mini-mock test** and answering timed with the timer on. When the time is over, don't stop answering. Just mark the point up to which you have answered in the scheduled time and go on to complete the test.

After the test score yourself on your answer in scheduled time and analyze all the difficulties by going through the corresponding solution set (link given at the end).

Now set the timer on and start taking the test.

### 12th question set- 10 problems for SSC CGL Tier II exam: 3rd on Trigonometry - testing time 12 mins

#### Problem 1.

If $2abcos \theta + (a^2-b^2)sin \theta=a^2+b^2$ 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 2.

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

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

#### Problem 3.

If $cos \theta + sec \theta = 2$, then the value of $cos^5 \theta + sec^5 \theta$ is,

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

#### Problem 4.

$sin(\alpha + \beta -\gamma)=cos(\beta + \gamma -\alpha)=\displaystyle\frac{1}{2}$ and $tan(\gamma + \alpha -\beta)=1$. If $\alpha$, $\beta$ and $\gamma$ are positive acute angles, value of $2\alpha + \beta$ is,

- $105^0$
- $110^0$
- $115^0$
- $120^0$

#### Problem 5.

If $sin \theta + sin^2 \theta=1$, then the value of $cos^{12} \theta + 3cos^{10} \theta + 3cos^{8} \theta + cos^6 \theta - 1$ is,

- $1$
- $0$
- $2$
- $3$

#### Problem 6.

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 7.

If $4cos^2 \theta - 4\sqrt{3}cos \theta + 3=0$ and $0^0 \leq \theta \leq 90^0$, then the value of $\theta$ is,

- $60^0$
- $90^0$
- $30^0$
- $45^0$

#### Problem 8.

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

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

#### Problem 9.

If $x=asin \theta-bcos \theta$ and $y=acos \theta + bsin \theta$, then which of the following is true?

- $x^2+y^2=a^2+b^2$
- $\displaystyle\frac{x^2}{a^2}+ \displaystyle\frac{y^2}{b^2}=1$
- $x^2+y^2=a^2-b^2$
- $\displaystyle\frac{x^2}{y^2}+ \displaystyle\frac{a^2}{b^2}=1$

#### Problem 10.

If $tan \theta=\displaystyle\frac{a}{b}$, then the value of $\displaystyle\frac{asin^3 \theta - bcos^3 \theta}{asin^3 \theta + bcos^3 \theta}$ is,

- $\displaystyle\frac{a^4-b^4}{a^4+b^4}$
- $\displaystyle\frac{a^3+b^3}{a^3-b^3}$
- $\displaystyle\frac{a^3-b^3}{a^3+b^3}$
- $\displaystyle\frac{a^4+b^4}{a^4-b^4}$

For detailed solutions refer to the companion **SSC CGL Tier II Solution set 12 Trigonometry 3, questions with solutions.**

You may also watch the video solutions at the two-part video below.

**Part 1: Q1 to Q5**

**Part 2: Q6 to Q10**

The answers to the questions are given below.

### Answers to the questions.

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

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

**Problem 3. Answer:** Option b: $2$.

**Problem 4. Answer:** Option d: $120^0$.

**Problem 5. Answer:** Option b: $0$.

**Problem 6. Answer:** Option c: 2.

**Problem 7. Answer:** Option c: $30^0$.

**Problem 8. Answer:** Option b: $\sqrt{2}+1$.

**Problem 9. Answer:** Option a: $x^2 + y^2=a^2 + b^2$.

**Problem 10. Answer:** Option a: $\displaystyle\frac{a^4-b^4}{a^4+b^4}$.

### Guided help on Trigonometry in Suresolv

To get the best results out of the extensive range of articles of **tutorials**, **questions** and **solutions** on **Trigonometry **in Suresolv, *follow the guide,*

**The guide list of articles is up-to-date.**