11th SSC CGL Tier II level Question Set, topic Trigonometry 2
This is the 11th question set of 10 practice problem exercise for SSC CGL Tier II level exam and 2nd on topic Trigonometry.
The answers to the questions and link to the 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.
11th question 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 $\sin \theta$,
- $-\displaystyle\frac{12}{13}$
- $\displaystyle\frac{12}{13}$
- $\displaystyle\frac{5}{13}$
- $\displaystyle\frac{6}{13}$
Problem 2.
The value of $(cosec \theta -sin \theta)(sec \theta - cos \theta)(tan \theta +cot \theta)$ is,
- 1
- 2
- 4
- 6
Problem 3.
If $tan A = n tan B$ and $sin A = m sin B$, then the value of $cos^2 A$ is,
- $\displaystyle\frac{m^2+1}{n^2-1}$
- $\displaystyle\frac{m^2+1}{n^2+1}$
- $\displaystyle\frac{m^2 -1}{n^2+1}$
- $\displaystyle\frac{m^2 -1}{n^2-1}$
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,
- $\displaystyle\frac{1}{\sqrt{2}}$
- $1$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{\sqrt{3}}{2}$
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,
- $-\displaystyle\frac{3}{5}$
- $-\displaystyle\frac{15}{9}$
- $\displaystyle\frac{17}{5}$
- $\displaystyle\frac{3}{5}$
Problem 6.
If $\sin (\theta + 30^0)=\displaystyle\frac{3}{\sqrt{2}}$ the value of $cos^2 \theta$ is,
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{1}{4}$
- $\displaystyle\frac{3}{4}$
- $\displaystyle\frac{\sqrt{3}}{2}$
Problem 7.
$(1 + sec 20^0 + cot 70^0)(1 - cosec 20^0 + tan 70^0)$ is equal to,
- $1$
- $0$
- $-1$
- $2$
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,
- $3$
- $\displaystyle\frac{1}{\sqrt{3}}$
- $\sqrt{3}$
- $\displaystyle\frac{1}{3}$
Problem 9.
If $sec \theta - tan \theta=\displaystyle\frac{1}{\sqrt{3}}$, then the value of $sec \theta.tan \theta$ is,
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{4}{\sqrt{3}}$
- $\displaystyle\frac{1}{\sqrt{3}}$
- $\displaystyle\frac{2}{\sqrt{3}}$
Problem 10.
If $tan (5x - 10^0)=cot (5y+20^0)$, then the value of $x+y$ is,
- $15^0$
- $16^0$
- $20^0$
- $24^0$
Answers to the problems
Problem 1. Answer: b: $\displaystyle\frac{12}{13}$.
Problem 2. Answer: a: 1.
Problem 3. Answer: d: $\displaystyle\frac{m^2-1}{n^2-1}$.
Problem 4. Answer: c: $\displaystyle\frac{1}{2}$
Problem 5. Answer: a: $-\displaystyle\frac{3}{5}$.
Problem 6. Answer: c: $\displaystyle\frac{3}{4}$.
Problem 7. Answer: d: 2.
Problem 8. Answer: a: 3.
Problem 9. Answer: a: $\displaystyle\frac{2}{3}$.
Problem 10. Answer: b: $16^0$.
For detailed conceptual solutions please refer to companion SSC CGL Tier II level solution set 11 Trigonometry 2.
You may watch the video solutions in the two-part video below.
Part 1: Q1 to Q5
Part 2: Q6 to Q10
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