Advanced Trigonometry: MCQ test, Set 10

Download the PDF (with key sheet and solutions, 92.5 KB)

Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

$\displaystyle \frac{1}{1 + \sin\theta} + \frac{1}{1 - \sin\theta} =$
A $2$ B $\operatorname{sec}^2\theta$ C $2\operatorname{csc}^2\theta$ D $2\operatorname{sec}^2\theta$

Answer and solution

Answer: (D) $2\operatorname{sec}^2\theta$

$\displaystyle = \frac{(1 - \sin\theta) + (1 + \sin\theta)}{1 - \sin^2\theta} = \frac{2}{\cos^2\theta} = 2\sec^2\theta$
The radian measure of $105^\circ$ is
A $\displaystyle \frac{5\pi}{12}$ B $\displaystyle \frac{7\pi}{12}$ C $\displaystyle \frac{11\pi}{12}$ D $\displaystyle \frac{7\pi}{6}$

Answer and solution

Answer: (B) $\displaystyle \frac{7\pi}{12}$

$\displaystyle 105^\circ = 105 \times \frac{\pi}{180} = \frac{7\pi}{12}$
If $\theta = 160^\circ$ , which of the following is positive?
A $\operatorname{tan}\theta$ B $\operatorname{cot}\theta$ C $\operatorname{cos}\theta$ D $\operatorname{csc}\theta$

Answer and solution

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

$160^\circ \in \text{II}_Q$ , where only sin, csc are +ve
$3\cot^2 60^\circ + \sec^2 45^\circ =$
A $2$ B $3$ C $\displaystyle \frac{3}{2}$ D $11$

Answer and solution

Answer: (B) $3$

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

$= 3$
$\operatorname{sec}(360^\circ - \theta) =$
A $\operatorname{sec}\theta$ B $-\operatorname{csc}\theta$ C $-\operatorname{sec}\theta$ D $\operatorname{csc}\theta$

Answer and solution

Answer: (A) $\operatorname{sec}\theta$

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

For acute $\theta$ , $360^\circ - \theta \in \text{IV}_Q$ , where $\sec$ is +ve

$\therefore \sec(360^\circ - \theta) = \sec\theta$
$\tan 1260^\circ + \cos 1170^\circ + \sin 1350^\circ =$
A $1$ B $0$ C $-2$ D $-1$

Answer and solution

Answer: (D) $-1$

$\tan 1260^\circ = \tan(1080^\circ + 180^\circ) = \tan 180^\circ = 0$

$\cos 1170^\circ = \cos(1080^\circ + 90^\circ) = \cos 90^\circ = 0$

$\sin 1350^\circ = \sin(1080^\circ + 270^\circ) = \sin 270^\circ = -1$

$0 + 0 + (-1)$

$= -1$
Which of the following is a possible value of $\cos\theta$ ?
A $\displaystyle \frac{7}{5}$ B $\displaystyle -\frac{5}{7}$ C $-2$ D $\sqrt{3}$

Answer and solution

Answer: (B) $\displaystyle -\frac{5}{7}$

$-1 \le \cos\theta \le 1$

Only $\displaystyle -\frac{5}{7}$ lies in $[-1, 1]$
Between $0$ and $2\pi$ , the graph of $\cos x$ cuts the x-axis at
A $\displaystyle \frac{\pi}{2} \text{ and } \frac{3\pi}{2}$ B $\pi \text{ and } 2\pi$ C $0 \text{ and } \pi$ D $\displaystyle \frac{\pi}{4} \text{ and } \frac{3\pi}{4}$

Answer and solution

Answer: (A) $\displaystyle \frac{\pi}{2} \text{ and } \frac{3\pi}{2}$

$\displaystyle \cos x = 0 \Rightarrow x = (2n + 1)\frac{\pi}{2}$

In $(0, 2\pi)$ : $\displaystyle x = \frac{\pi}{2}, \frac{3\pi}{2}$
$\cot 12^\circ \cot 38^\circ \cot 52^\circ \cot 60^\circ \cot 78^\circ =$
A $\sqrt{3}$ B $0$ C $\displaystyle \frac{1}{\sqrt{3}}$ D $1$

Answer and solution

Answer: (C) $\displaystyle \frac{1}{\sqrt{3}}$

$\cot 12^\circ \cot 78^\circ = 1$ , $\cot 38^\circ \cot 52^\circ = 1$

G.E $\displaystyle = \cot 60^\circ = \frac{1}{\sqrt{3}}$
If $\displaystyle \sin\theta\cos\theta = \frac{1}{8}$ , then $\sin^4\theta + \cos^4\theta =$
A $\displaystyle \frac{31}{32}$ B $\displaystyle \frac{61}{64}$ C $\displaystyle \frac{63}{64}$ D $\displaystyle \frac{15}{16}$

Answer and solution

Answer: (A) $\displaystyle \frac{31}{32}$

$\sin^4\theta + \cos^4\theta = (\sin^2\theta + \cos^2\theta)^2 - 2\sin^2\theta\cos^2\theta = 1 - 2(\sin\theta\cos\theta)^2$

$\displaystyle = 1 - 2\left(\frac{1}{8}\right)^2 = 1 - \frac{1}{32} = \frac{31}{32}$
If $x > 0$ , the least value of $\displaystyle x + \frac{4}{x}$ is
A $4$ B $8$ C $2$ D $5$

Answer and solution

Answer: (A) $4$

A.M $\ge$ G.M: $\displaystyle \frac{x + \frac{4}{x}}{2} \ge \sqrt{x \cdot \frac{4}{x}} = 2$

$\displaystyle \Rightarrow x + \frac{4}{x} \ge 4$ (equality at $x = 2$ )
$\sin(45^\circ + \theta) - \cos(45^\circ - \theta) =$
A $\sqrt{2}\operatorname{cos}\theta$ B $2\operatorname{sin}\theta$ C $0$ D $\sqrt{2}\operatorname{sin}\theta$

Answer and solution

Answer: (C) $0$

$\displaystyle \sin(45^\circ + \theta) = \frac{1}{\sqrt{2}}(\cos\theta + \sin\theta)$

$\displaystyle \cos(45^\circ - \theta) = \frac{1}{\sqrt{2}}(\cos\theta + \sin\theta)$

Difference $= 0$
$\tan 80^\circ - \tan 10^\circ =$
A $2\operatorname{cot} 70^\circ$ B $\operatorname{tan} 70^\circ$ C $2\operatorname{tan} 70^\circ$ D $2\operatorname{tan} 10^\circ$

Answer and solution

Answer: (C) $2\operatorname{tan} 70^\circ$

$\displaystyle \tan 70^\circ = \tan(80^\circ - 10^\circ) = \frac{\tan 80^\circ - \tan 10^\circ}{1 + \tan 80^\circ\tan 10^\circ}$

$\tan 80^\circ\tan 10^\circ = 1$ , so $\tan 80^\circ - \tan 10^\circ = 2\tan 70^\circ$
If $\displaystyle \cos A = \frac{3}{4}$ and $\displaystyle \sin B = \frac{1}{2}$ , then $\cos(A + B)\cos(A - B) =$
A $\displaystyle -\frac{3}{16}$ B $\displaystyle \frac{1}{4}$ C $\displaystyle \frac{13}{16}$ D $\displaystyle \frac{5}{16}$

Answer and solution

Answer: (D) $\displaystyle \frac{5}{16}$

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

$\displaystyle = \frac{9}{16} - \frac{1}{4} = \frac{5}{16}$
If $A + B + C = 90^\circ$ , $\cot A = 2$ and $\cot B = 3$ , then $\cot C =$
A $5$ B $-1$ C $1$ D $\displaystyle \frac{1}{5}$

Answer and solution

Answer: (C) $1$

$\Sigma\cot A = \Pi\cot A$ : $2 + 3 + \cot C = 6\cot C$

$5\cot C = 5 \Rightarrow \cot C = 1$
$(2 + 2\tan 15^\circ)(1 + \tan 30^\circ) =$
A $8$ B $4$ C $2$ D $3$

Answer and solution

Answer: (B) $4$

$= 2(1 + \tan 15^\circ)(1 + \tan 30^\circ)$ and $15^\circ + 30^\circ = 45^\circ$

$= 2 \times 2 = 4$
$8\sin x\cos x\cos 2x\cos 4x =$
A $\operatorname{sin} 16x$ B $8\operatorname{sin} 8x$ C $\operatorname{sin} 4x$ D $\operatorname{sin} 8x$

Answer and solution

Answer: (D) $\operatorname{sin} 8x$

$= 4\sin 2x\cos 2x\cos 4x = 2\sin 4x\cos 4x = \sin 8x$
If $\displaystyle \tan\frac{A}{2} = 2$ , then $\cos A =$
A $\displaystyle -\frac{3}{5}$ B $\displaystyle \frac{4}{5}$ C $\displaystyle -\frac{4}{3}$ D $\displaystyle \frac{3}{5}$

Answer and solution

Answer: (A) $\displaystyle -\frac{3}{5}$

$\displaystyle \cos A = \frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}} = \frac{1 - 4}{1 + 4} = -\frac{3}{5}$
$\displaystyle \tan 7\tfrac{1}{2}^\circ =$
A $(\sqrt{3} - \sqrt{2})(\sqrt{2} - 1)$ B $(\sqrt{3} - \sqrt{2})(\sqrt{2} + 1)$ C $(\sqrt{3} + \sqrt{2})(\sqrt{2} + 1)$ D $(\sqrt{3} + \sqrt{2})(\sqrt{2} - 1)$

Answer and solution

Answer: (A) $(\sqrt{3} - \sqrt{2})(\sqrt{2} - 1)$

$\displaystyle \tan 7\tfrac{1}{2}^\circ = \frac{1 - \cos 15^\circ}{\sin 15^\circ}$

Simplifying gives $(\sqrt{3} - \sqrt{2})(\sqrt{2} - 1)$ (standard result)
$\displaystyle \cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7} =$
A $\displaystyle \frac{1}{16}$ B $\displaystyle -\frac{1}{8}$ C $\displaystyle \frac{1}{8}$ D $\displaystyle \frac{1}{4}$

Answer and solution

Answer: (C) $\displaystyle \frac{1}{8}$

$\displaystyle = \frac{\sin\frac{16\pi}{7}}{8\sin\frac{2\pi}{7}}$

$\displaystyle \sin\frac{16\pi}{7} = \sin\left(2\pi + \frac{2\pi}{7}\right) = \sin\frac{2\pi}{7}$

G.E $\displaystyle = \frac{1}{8}$
If $\displaystyle a = \cos\frac{2\pi}{7}$ , $\displaystyle b = \cos\frac{4\pi}{7}$ , $\displaystyle c = \cos\frac{6\pi}{7}$ , then $8abc =$
A $-1$ B $4$ C $-4$ D $1$

Answer and solution

Answer: (D) $1$

$\displaystyle abc = -\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7} = -\frac{\sin\frac{8\pi}{7}}{8\sin\frac{\pi}{7}} = \frac{1}{8}$

$8abc = 1$
$\tan 5^\circ\tan 55^\circ\tan 65^\circ =$
A $\operatorname{tan} 5^\circ$ B $2 - \sqrt{3}$ C $2 + \sqrt{3}$ D $\displaystyle \frac{1}{\sqrt{3}}$

Answer and solution

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

$\theta = 5^\circ$ : $= \tan 15^\circ = 2 - \sqrt{3}$
$\sqrt{1 + \sin 240^\circ} =$
A $\displaystyle \frac{\sqrt{3} + 1}{2}$ B $\displaystyle \frac{1 - \sqrt{3}}{2}$ C $\displaystyle \frac{\sqrt{3} - 1}{2}$ D $\displaystyle \frac{\sqrt{3}}{2}$

Answer and solution

Answer: (C) $\displaystyle \frac{\sqrt{3} - 1}{2}$

$\sqrt{1 + \sin 2A} = |\cos A + \sin A|$ with $A = 120^\circ$

$\displaystyle = \left|-\frac{1}{2} + \frac{\sqrt{3}}{2}\right| = \frac{\sqrt{3} - 1}{2}$
If $\displaystyle \sin x + \sin y = \frac{1}{2}$ and $\displaystyle \cos x + \cos y = \frac{1}{3}$ , then $\displaystyle \tan\frac{x + y}{2} =$
A $\displaystyle \frac{2}{3}$ B $\displaystyle \frac{5}{6}$ C $\displaystyle \frac{3}{2}$ D $\displaystyle \frac{1}{6}$

Answer and solution

Answer: (C) $\displaystyle \frac{3}{2}$

$\displaystyle \frac{\sin x + \sin y}{\cos x + \cos y} = \frac{2\sin\frac{x + y}{2}\cos\frac{x - y}{2}}{2\cos\frac{x + y}{2}\cos\frac{x - y}{2}} = \tan\frac{x + y}{2}$

$\displaystyle = \frac{1/2}{1/3} = \frac{3}{2}$
The period of $\tan^3 x$ is
A $2\pi$ B $\pi$ C $3\pi$ D $\displaystyle \frac{\pi}{3}$

Answer and solution

Answer: (B) $\pi$

$\tan^n x$ has period $\pi$ for every $n$

Menu

MCQ test, Set 10