SSC CGL Tier 2 Question Set 11 on Trigonometry with answers
Solve 10 selected questions in SSC CGL Tier 2 trigonometry questions set 11 in 12 minutes. Verify results from given answers. Learn to solve from solutions.
Answers and linked solution set are at the end.
SSC CGL Tier 2 Trigonometry Questions set 11 - testing time 12 mins
Problem 1.
If $5 cos \theta +12 sin \theta=13$, and $0^0 \lt \theta \lt 90^0$, then the value of $\sin \theta$,
- $-\displaystyle\frac{12}{13}$
- $\displaystyle\frac{12}{13}$
- $\displaystyle\frac{5}{13}$
- $\displaystyle\frac{6}{13}$
Problem 2.
The value of $(cosec \theta -sin \theta)(sec \theta - cos \theta)(tan \theta +cot \theta)$ is,
- 1
- 2
- 4
- 6
Problem 3.
If $tan A = n tan B$ and $sin A = m sin B$, then the value of $cos^2 A$ is,
- $\displaystyle\frac{m^2+1}{n^2-1}$
- $\displaystyle\frac{m^2+1}{n^2+1}$
- $\displaystyle\frac{m^2 -1}{n^2+1}$
- $\displaystyle\frac{m^2 -1}{n^2-1}$
Problem 4.
If $\theta$ is a positive acute angle and $3(sec^2 \theta + tan^2 \theta)=5$, then the value of $cos 2\theta$ is,
- $\displaystyle\frac{1}{\sqrt{2}}$
- $1$
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{\sqrt{3}}{2}$
Problem 5.
If $tan \alpha = 2$, then the value of $\displaystyle\frac{cosec^2 \alpha - sec^2 \alpha}{cosec^2+sec^2 \alpha}$ is,
- $-\displaystyle\frac{3}{5}$
- $-\displaystyle\frac{15}{9}$
- $\displaystyle\frac{17}{5}$
- $\displaystyle\frac{3}{5}$
Problem 6.
If $\sin (\theta + 30^0)=\displaystyle\frac{3}{\sqrt{2}}$ the value of $cos^2 \theta$ is,
- $\displaystyle\frac{1}{2}$
- $\displaystyle\frac{1}{4}$
- $\displaystyle\frac{3}{4}$
- $\displaystyle\frac{\sqrt{3}}{2}$
Problem 7.
$(1 + sec 20^0 + cot 70^0)(1 - cosec 20^0 + tan 70^0)$ is equal to,
- $1$
- $0$
- $-1$
- $2$
Problem 8.
If $tan \theta - cot \theta =0$, and $\theta$ is a positive acute angle, then the value of $\displaystyle\frac{tan (\theta+15^0)}{tan(\theta-15^0)}$ is,
- $3$
- $\displaystyle\frac{1}{\sqrt{3}}$
- $\sqrt{3}$
- $\displaystyle\frac{1}{3}$
Problem 9.
If $sec \theta - tan \theta=\displaystyle\frac{1}{\sqrt{3}}$, then the value of $sec \theta.tan \theta$ is,
- $\displaystyle\frac{2}{3}$
- $\displaystyle\frac{4}{\sqrt{3}}$
- $\displaystyle\frac{1}{\sqrt{3}}$
- $\displaystyle\frac{2}{\sqrt{3}}$
Problem 10.
If $tan (5x - 10^0)=cot (5y+20^0)$, then the value of $x+y$ is,
- $15^0$
- $16^0$
- $20^0$
- $24^0$
Answers to SSC CGL Tier 2 Trigonometry Questions set 11
Problem 1. Answer: b: $\displaystyle\frac{12}{13}$.
Problem 2. Answer: a: 1.
Problem 3. Answer: d: $\displaystyle\frac{m^2-1}{n^2-1}$.
Problem 4. Answer: c: $\displaystyle\frac{1}{2}$
Problem 5. Answer: a: $-\displaystyle\frac{3}{5}$.
Problem 6. Answer: c: $\displaystyle\frac{3}{4}$.
Problem 7. Answer: d: 2.
Problem 8. Answer: a: 3.
Problem 9. Answer: a: $\displaystyle\frac{2}{3}$.
Problem 10. Answer: b: $16^0$.
For detailed conceptual solutions please refer to companion SSC CGL Tier II level solution set 11 Trigonometry 2.
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