SSC CGL Tier II level Question Set 11, Trigonometry 2

Submitted by Atanu Chaudhuri on Sat, 26/05/2018 - 23:44

11th SSC CGL Tier II level Question Set, topic Trigonometry 2

SSC CGL tier 2 question set 11 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.

We repeat the method of taking the test. It is important to follow result bearing methods even in practice test environment.

Method of taking the test for getting the best results from the test:

Before start, you may refer to our tutorial Basic and rich Trigonometric concepts and applications or any short but good material to refresh your concepts if you so require.
Answer the questions in an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes.
When the time limit of 12 minutes is over, mark up to which you have answered, but go on to complete the set.
At the end, refer to the answers given at the end to mark your score at 12 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time.
Identify and analyze the problems that you couldn't do to learn how to solve those problems.
Identify and analyze the problems that you solved incorrectly. Identify the reasons behind the errors. If it is because of your shortcoming in topic knowledge improve it by referring to only that part of concept from the best source you can get hold of. You might google it. If it is because of your method of answering, analyze and improve those aspects specifically.
Identify and analyze the problems that posed difficulties for you and delayed you. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques.
Give a gap before you take a 10 problem practice test again.

Important: both practice tests and mock tests must be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance.

Resources that should be useful for you

You may refer to:

7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests or section on SSC CGL to access all the valuable student resources that we have created specifically for SSC CGL, but generally for any hard MCQ test.

If you like, you may subscribe to get latest content from this place.

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,

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 $\displaystyle\frac{cosec^2 \alpha - sec^2 \alpha}{cosec^2+sec^2 \alpha}$ 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.

Resources on Trigonometry and related topics

You may refer to our useful resources on Trigonometry and other related topics especially algebra.

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.