SSC CGL level Question Set 16, Trigonometry

Submitted by Atanu Chaudhuri on Thu, 12/05/2016 - 17:22

Sixteenth SSC CGL level Question Set on Trigonometry

ssc cgl question set 16 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 of $(sec\theta - cos\theta)^2 + (cosec\theta - sin\theta)^2 - (cot\theta - tan\theta)^2$ 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,

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.

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.