MCQ test, Set 1

Answer: (B) $\displaystyle \frac{12}{5}$

$\displaystyle \sec\theta = \frac{\text{hyp}}{\text{adj}} = \frac{13}{5}$: take hyp $= 13$, adj $= 5$

opp $= \sqrt{13^2 - 5^2} = \sqrt{144} = 12$

$\displaystyle \tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{12}{5}$

Answer: (A) $\displaystyle \frac{7\pi}{6}$

$\displaystyle 210^\circ = 210 \times \frac{\pi}{180} = \frac{7\pi}{6}$

Answer: (D) $\operatorname{tan}\theta$

$230^\circ \in \text{III}_Q$, where only tan, cot are +ve

Answer: (C) $\displaystyle \frac{1}{2}$

$\displaystyle = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{1}{2}\cdot\frac{1}{2} = \frac{3}{4} - \frac{1}{4}$

$\displaystyle = \frac{1}{2}$

Answer: (C) $-\operatorname{cot}\theta$

$180^\circ - \theta$ is of the form $180^\circ \pm \theta$ or $360^\circ \pm \theta$, so $\cot$ does not change

For acute $\theta$, $180^\circ - \theta \in \text{II}_Q$, where $\cot$ is -ve

$\therefore \cot(180^\circ - \theta) = -\cot\theta$

Answer: (D) $\displaystyle -\frac{\sqrt{3}}{2}$

$\cos 1290^\circ = \cos(1080^\circ + 210^\circ) = \cos 210^\circ$

$210^\circ \in \text{III}_Q$, where $\cos$ is -ve

$\displaystyle \cos 210^\circ = \cos(180^\circ + 30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$

Answer: (B) $(-\infty, -1] \cup [1, \infty)$

$\displaystyle \sec x = \frac{1}{\cos x}$ and $-1 \le \cos x \le 1$, $\cos x \ne 0$

So $\sec x \le -1$ or $\sec x \ge 1$

Answer: (A) $\operatorname{sin} 2 > \operatorname{sin} 1 > \operatorname{sin} 3$

$1 \approx 57^\circ$, $2 \approx 114^\circ$, $3 \approx 171^\circ$

$\sin 2 = \sin 66^\circ$, $\sin 3 = \sin 9^\circ$

$\sin x$ increases on $(0^\circ, 90^\circ)$: $\sin 66^\circ > \sin 57^\circ > \sin 9^\circ$

Answer: (C) $1$

$\cos 53^\circ = \cos(90^\circ - 37^\circ) = \sin 37^\circ$

So the ratio is $1$

Answer: (D) $1$

$\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta$

Adding $3\sin^2\theta\cos^2\theta$ gives $1$

Answer: (A) $2$

Consider $\cos^2\theta$ and $\sec^2\theta$; both are +ve

A.M $\ge$ G.M: $\displaystyle \frac{\cos^2\theta + \sec^2\theta}{2} \ge \sqrt{\cos^2\theta \cdot \sec^2\theta} = \sqrt{1} = 1$

$\Rightarrow \cos^2\theta + \sec^2\theta \ge 2$; equality when $\cos^2\theta = 1$

Answer: (D) $\displaystyle \frac{\sqrt{3}+1}{2\sqrt{2}}$

$\sin(45^\circ + 30^\circ) = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ$

$\displaystyle = \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\cdot\frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

Answer: (B) $2 + \sqrt{3}$

$\displaystyle \tan(45^\circ + 30^\circ) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$

$\displaystyle = \frac{(\sqrt{3} + 1)^2}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$

Answer: (C) $\displaystyle \frac{1}{\sqrt{2}}\operatorname{sin}\theta$

$\displaystyle = \sin\left(\frac{\pi}{4}\right)\sin\theta = \frac{1}{\sqrt{2}}\sin\theta$

Answer: (C) $\operatorname{tan} A\operatorname{tan} B\operatorname{tan} C$

$\displaystyle A + B = 180^\circ - C \Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan B} = -\tan C$

$\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C$

Answer: (D) $2$

$8^\circ + 37^\circ = 45^\circ \Rightarrow (1 + \tan A)(1 + \tan B) = 2$

Answer: (A) $\displaystyle \frac{1}{\sqrt{2}}$

$2\sin A\cos A = \sin 2A$

$\displaystyle = \sin 45^\circ = \frac{1}{\sqrt{2}}$

Answer: (C) $\displaystyle \frac{11}{16}$

$\displaystyle \sin 3A = 3\sin A - 4\sin^3 A = \frac{3}{4} - 4\cdot\frac{1}{64} = \frac{3}{4} - \frac{1}{16} = \frac{11}{16}$

Answer: (D) $\displaystyle \frac{\sqrt{5}-1}{4}$

Standard value: $\displaystyle \sin 18^\circ = \frac{\sqrt{5}-1}{4}$

Answer: (A) $2\sqrt{3}$

$\cot A - \tan A = 2\cot 2A$

$= 2\cot 30^\circ = 2\sqrt{3}$

Answer: (D) $\displaystyle \frac{1}{2}$

$\displaystyle a_1 = \frac{\pi}{7}$, $\displaystyle d = \frac{2\pi}{7}$, $n = 3$: sum $\displaystyle = \frac{\cos\frac{3\pi}{7}\sin\frac{3\pi}{7}}{\sin\frac{\pi}{7}} = \frac{\sin\frac{6\pi}{7}}{2\sin\frac{\pi}{7}} = \frac{1}{2}$

Answer: (B) $\displaystyle \frac{\sqrt{3}}{8}$

$\displaystyle \sin\theta\sin(60^\circ - \theta)\sin(60^\circ + \theta) = \frac{1}{4}\sin 3\theta$ with $\theta = 20^\circ$

$\displaystyle = \frac{1}{4}\sin 60^\circ = \frac{\sqrt{3}}{8}$

Answer: (B) $2\operatorname{sin} 25^\circ$

$\sqrt{1 + \sin 50^\circ} = \cos 25^\circ + \sin 25^\circ$, $\sqrt{1 - \sin 50^\circ} = \cos 25^\circ - \sin 25^\circ$ ($25^\circ < 45^\circ$)

Difference $= 2\sin 25^\circ$

Answer: (A) $\displaystyle \frac{2 + \sqrt{3}}{2}$

$2\sin A\cos B = \sin(A + B) + \sin(A - B)$

$\displaystyle = \sin 90^\circ + \sin 60^\circ = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$

Answer: (B) $\displaystyle \frac{2\pi}{5}$

Period of $\cos(ax + b) + k$ is $\displaystyle \frac{2\pi}{|a|} = \frac{2\pi}{5}$