Hard trigonometry questions with answers for competitive exams SSC CGL Set 65
10 hard trigonometry questions with answers for SSC CGL. Take the timed test, verify from answers and learn to solve quickly from solutions (link given).
This is the 7th set of 10 questions on Trigonometry for competitive exams SSC CGL.
Answers and link to the solutions are at the end.
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Hard trigonometry questions with answers for competitive exams SSC CGL Set 65 - time to solve 12 mins
Problem 1.
If $2-cos^2 \theta=3sin \theta{cos \theta}$, where $sin \theta \neq cos \theta$, the value of $tan \theta$ is,
- $0$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{1}{3}$
Problem 2.
If $sin \theta + cos \theta =\sqrt{2}cos(90^0- \theta)$, then $cot \theta$ is,
- $\sqrt{2}-1$
- $\sqrt{2}+1$
- $0$
- $\sqrt{2}$
Problem 3.
If $(a^2-b^2)sin \theta + 2abcos \theta=a^2+b^2$ then the value of $tan \theta$ is,
- $\displaystyle\frac{1}{2ab}(a^2+b^2)$
- $\displaystyle\frac{1}{2}(a^2-b^2)$
- $\displaystyle\frac{1}{2}(a^2+b^2)$
- $\displaystyle\frac{1}{2ab}(a^2-b^2)$
Problem 4.
The value of $sec \theta\left(\displaystyle\frac{1+sin \theta}{cos \theta}+\displaystyle\frac{cos \theta}{1+sin \theta}\right) - 2tan^2 \theta$ is,
- 4
- 0
- 2
- 1
Problem 5.
If $cot \theta + cosec \theta =3$, and $\theta$ an acute angle, the value of $cos \theta$ is,
- $1$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{3}{4}$
Problem 6.
If $xcos \theta - sin \theta=1$, then the value of $x^2-(1+x^2)sin \theta$ is,
- $1$
- $0$
- $2$
- $-1$
Problem 7.
If $\theta=60^0$, then the value of $\displaystyle\frac{1}{2}\sqrt{1+ sin \theta} + \displaystyle\frac{1}{2}\sqrt{1- sin \theta}$ is,
- $cot \displaystyle\frac{\theta}{2}$
- $cos \displaystyle\frac{\theta}{2}$
- $sec \displaystyle\frac{\theta}{2}$
- $sin \displaystyle\frac{\theta}{2}$
Problem 8.
If $3sin \theta + 5cos \theta =5$, ($0\lt \theta \lt 90^0$), then the value of $5sin \theta-3cos \theta$ will be,
- 1
- 2
- 5
- 3
Problem 9.
If $tan \theta = \displaystyle\frac{1}{\sqrt{11}}$, and $0\lt \theta \lt 90^0$, then the value of $\displaystyle\frac{cosec^2 \theta - sec^2 \theta}{cosec^2 \theta + sec^2 \theta}$ is,
- $\displaystyle\frac{5}{6}$
- $\displaystyle\frac{3}{4}$
- $\displaystyle\frac{4}{5}$
- $\displaystyle\frac{6}{7}$
Problem 10.
If $tan^2 \theta=1-e^2$, then the value of $sec \theta + tan^3 \theta{cosec \theta}$ is equal to,
- $(2+e^2)^{\frac{3}{2}}$
- $(2+e^2)^{\frac{1}{2}}$
- $(2-e^2)^{\frac{3}{2}}$
- $(2-e^2)^{\frac{1}{2}}$
The answers are given below.
Learn how to solve the questions in 12 minutes' scheduled time from the paired solution set,
SSC CGL level Solution Set 65 on Trigonometry 6.
Answers to the 10 hard trigonometry questions for competitive exams SSC CGL Set 65
Problem 1. Answer: b: $\displaystyle\frac{1}{2}$.
Problem 2. Answer: a: $\sqrt{2}-1$.
Problem 3. Answer: d: $\displaystyle\frac{1}{2ab}(a^2-b^2)$.
Problem 4. Answer: Option c: 2.
Problem 5. Answer: c: $\displaystyle\frac{4}{5}$.
Problem 6. Answer: a: $1$.
Problem 7. Answer: b: $cos \displaystyle\frac{\theta}{2}$.
Problem 8. Answer: d: 3.
Problem 9. Answer: a: $\displaystyle\frac{5}{6}$.
Problem 10. Answer: c: $(2-e^2)^{\frac{3}{2}}$.
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