SSC CGL level Question Set 19, Trigonometry 3 | SureSolv

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SSC CGL level Question Set 19, Trigonometry 3

Nineteenth SSC CGL level Question Set, topic Trigonometry 3

SSC CGL Question Set 19 trigonometry 3

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

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.

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


Nineteenth question set- 10 problems for SSC CGL exam: 3rd on Trigonometry - time 12 mins

Problem 1.

The value of $tan1^0tan2^0tan3^0.....tan89^0$ is,

  1. $\sqrt{3}$
  2. $0$
  3. $1$
  4. $\displaystyle\frac{1}{\sqrt{3}}$

Problem 2.

The value of $cot18^0\left(cot72^0cos^222^0 + \displaystyle\frac{1}{tan72^0sec^268^0}\right)$ is,

  1. $\displaystyle\frac{1}{\sqrt{3}}$
  2. $3$
  3. $1$
  4. $\sqrt{2}$

Problem 3.

If $asin\theta + bcos\theta =c$, then the value of $acos\theta - bsin\theta$ is,

  1. $\pm \sqrt{-a^2 + b^2 + c^2}$
  2. $\pm \sqrt{a^2 - b^2 + c^2}$
  3. $\pm \sqrt{a^2 - b^2 - c^2}$
  4. $\pm \sqrt{a^2 + b^2 - c^2}$

Problem 4.

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

  1. 1
  2. 2
  3. 3
  4. 4

Problem 5.

$\displaystyle\frac{tan\theta}{1 - cot\theta} + \displaystyle\frac{cot\theta}{1 - tan\theta}$ is equal to,

  1. $1 - tan\theta -cot\theta$
  2. $1 + tan\theta + cot\theta$
  3. $1 - tan\theta + cot\theta$
  4. $1 + tan\theta - cot\theta$

Problem 6.

If $tan\theta = \displaystyle\frac{sin\alpha - cos\alpha}{sin\alpha + cos\alpha}$ then $sin\alpha + cos\alpha$ is,

  1. $\pm \sqrt{2} sin\theta$
  2. $\pm \sqrt{2} cos\theta$
  3. $\pm \displaystyle\frac{1}{\sqrt{2}} cos\theta$
  4. $\pm \displaystyle\frac{1}{\sqrt{2}} sin\theta$

Problem 7.

If $cos^2\alpha - sin^2\alpha = tan^2\beta$, then $cos^2\beta - sin^2\beta = $

  1. $tan^2\alpha$
  2. $cot^2\alpha$
  3. $cot^2\beta$
  4. $tan^2\beta$

Problem 8.

If $tan\alpha=ntan\beta$, and $sin\alpha = msin\beta$ then $cos^2\alpha$ is,

  1. $\displaystyle\frac{m^2 - 1}{n^2 - 1}$
  2. $\displaystyle\frac{m^2 + 1}{n^2 + 1}$
  3. $\displaystyle\frac{m^2}{n^2 + 1}$
  4. $\displaystyle\frac{m^2}{n^2}$

Problem 9.

If $A$, $B$ and $C$ are the three angles of a triangle, then the incorrect relation among the following is,

  1. $cos\displaystyle\frac{A + B}{2} = sin\displaystyle\frac{C}{2}$
  2. $sin\displaystyle\frac{A + B}{2} = cos\displaystyle\frac{C}{2}$
  3. $cot\displaystyle\frac{A + B}{2} = tan\displaystyle\frac{C}{2}$
  4. $tan\displaystyle\frac{A + B}{2} = sec\displaystyle\frac{C}{2}$

Problem 10.

If $\theta$ is a positive acute angle and $tan2\theta{tan3\theta} = 1$ then the value of $\left(2cos^2\displaystyle\frac{5\theta}{2} - 1\right)$ is,

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

You will find the detailed conceptual solutions to these questions in SSC CGL level Solution Set 19 on Trigonometry.

You may also watch the video solutions in the two-part video.

Part 1: Q1 to Q5

Part 2: Q6 to Q10

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.


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,

Reading and Practice Guide on Trigonometry in Suresolv for SSC CHSL, SSC CGL, SSC CGL Tier II Other Competitive exams.

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


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