Difficult Trigonometry questions with answers SSC CGL Set 40
Trigonometry high level questions for SSC CGL Set 40: 10 questions to solve in 12 min, answers verified and doubts clarified from paired solution set.
Answer to the questions and link to the paired solution set are at the end.
10 Trigonometry high level questions for SSC CGL Set 40 - test time 12 mins
Problem 1.
If $cosec 39^0=p$, the value of, $\displaystyle\frac{1}{cosec^2 51^0} + sin^2 39^0 + tan^2 51^0 - \displaystyle\frac{1}{sin^2 51^0 sec^2 39^0}$ is,
- $p^2 - 1$
- $\sqrt{p^2 - 1}$
- $1-p^2$
- $\sqrt{1-p^2}$
Problem 2.
If $sec \theta=x + \displaystyle\frac{1}{4x}$, where $(0^0 \lt \theta \lt 90^0)$, then $sec \theta + tan \theta$ is,
- $\displaystyle\frac{x}{2}$
- $2x$
- $\displaystyle\frac{2}{x}$
- $x$
Problem 3.
If $tan \theta=1$, then the value of $\displaystyle\frac{8sin \theta + 5cos \theta}{sin^3 \theta -2cos^3 \theta + 7cos \theta}$ is,
- $2\displaystyle\frac{1}{2}$
- $2$
- $3$
- $\displaystyle\frac{4}{5}$
Problem 4.
If $7sin \theta = 24cos \theta$, where $0 \lt \theta \lt \displaystyle\frac{\pi}{2}$, then the value of $14tan \theta - 75cos \theta - 7sec \theta$ is,
- 1
- 3
- 2
- 4
Problem 5.
The minimum value of $sin^2 \theta + cos^2 \theta + sec^2 \theta + cosec^2 \theta + tan^2 \theta + cot^2 \theta$ is equal to,
- 1
- 7
- 3
- 5
Problem 6.
In a right $\triangle ABC$ with right angle at $\angle ABC$, if $AB=2\sqrt{6}$ and $AC - BC = 2$ then, $sec A + tan A$ is,
- $\displaystyle\frac{\sqrt{6}}{2}$
- $2\sqrt{6}$
- $\sqrt{6}$
- $\displaystyle\frac{1}{\sqrt{6}}$
Problem 7.
If $tan 2\theta . tan 4\theta = 1$, then the value of $tan 3\theta$ is,
- $\sqrt {3}$
- $0$
- $\displaystyle\frac{1}{\sqrt{3}}$
- $1$
Problem 8.
If $sin \displaystyle\frac{\pi x}{2}=x^2 -2x +2$, then the value of $x$ is,
- $0$
- $-1$
- $1$
- none of these
Problem 9.
If $2sin \theta + cos \theta = \displaystyle\frac{7}{3}$, then the value of $(tan^2 \theta - sec^2 \theta)$ is,
- $0$
- $\displaystyle\frac{7}{3}$
- $\displaystyle\frac{3}{7}$
- $-1$
Problem 10.
If $(rcos \theta - \sqrt{3})^2 + (rsin \theta - 1)^2 = 0$, then the value of $\displaystyle\frac{rtan \theta + sec \theta}{rsec \theta + tan \theta}$ is,
- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{\sqrt{3}}{4}$
- $\displaystyle\frac{\sqrt{5}}{4}$
- $\displaystyle\frac{5}{4}$
Learn how to solve the questions easy and quick from the paired solution srt,
SSC CGL level Solution Set 40 on Trigonometry 4.
Answers to the Trigonometry high level questions for SSC CGL Set 40
Problem 1. Ans: Option a: $p^2 - 1$.
Problem 2. Ans: Option b: $2x$.
Problem 3. Ans: Option b: 2.
Problem 4. Ans: Option c: 2.
Problem 5. Ans: Option b: 7.
Problem 6. Ans: Option c: $\sqrt{6}$.
Problem 7. Ans: Option d: 1.
Problem 8. Ans: Option c: 1.
Problem 9. Ans: Option d: $-1$.
Problem 10. Ans: Option a: $\displaystyle\frac{4}{5}$.
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