SSC CGL level Question Set 19, Trigonometry 3

Trigonometry aptitude questions with answers

Solve 10 trigonometry aptitude questions for SSC CGL Set 19 in 12 minutes. Verify correctness from answers. Learn to solve quickly from paired solution set.

Answers and link to the solutions are at the end.

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10 Trigonometry aptitude questions for SSC CGL Set 19 - time to solve 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. $\displaystyle\frac{1}{2}$

Learn to solve the questions quickly from the paired solutions at,

SSC CGL level Solution Set 19 on Trigonometry.

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Answers to Trigonometry aptitude questions for SSC CGL Set 19

Problem 1. Answer: Option c: 1.

Problem 2. Answer: Option c: 1.

Problem 3. Answer: Option d: $\pm \sqrt{a^2 + b^2 - c^2}$.

Problem 4. Answer: Option b: 2.

Problem 5. Answer: Option b: $1 + tan\theta + cot\theta$.

Problem 6. Answer: Option b: $\pm \sqrt{2} cos\theta$.

Problem 7. Answer: Option a: $tan^2\alpha$.

Problem 8. Answer: Option a: $\displaystyle\frac{m^2 - 1}{n^2 - 1}$.

Problem 9. Answer: Option d: $tan\displaystyle\frac{A + B}{2} = sec\displaystyle\frac{C}{2}$.

Problem 10. Answer: Option a: $0$.

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