Advanced Trigonometry: MCQ test, Set 6

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

$\cot^2\theta - \csc^2\theta =$
A $0$ B $1$ C $-1$ D $2$

Answer and solution

Answer: (C) $-1$

$\csc^2\theta - \cot^2\theta = 1$

$\therefore \cot^2\theta - \csc^2\theta = -1$
$\displaystyle \frac{3\pi}{4}$ radians in grades is
A $150^g$ B $135^g$ C $75^g$ D $300^g$

Answer and solution

Answer: (A) $150^g$

$\pi = 200^g$

$\displaystyle \frac{3\pi}{4} = \frac{3}{4} \times 200^g = 150^g$
The sign of $\sec 250^\circ\cot 340^\circ$ is
A positive B zero C not defined D negative

Answer and solution

Answer: (A) positive

$250^\circ \in \text{III}_Q \Rightarrow \sec 250^\circ$ is -ve

$340^\circ \in \text{IV}_Q \Rightarrow \cot 340^\circ$ is -ve

So the product is positive
$\sin^2 45^\circ + \cos^2 60^\circ =$
A $1$ B $\displaystyle \frac{3}{4}$ C $\displaystyle \frac{1}{2}$ D $\displaystyle \frac{5}{4}$

Answer and solution

Answer: (B) $\displaystyle \frac{3}{4}$

$\displaystyle = \frac{1}{2} + \frac{1}{4}$

$\displaystyle = \frac{3}{4}$
$\operatorname{sec}(270^\circ - \theta) =$
A $\operatorname{csc}\theta$ B $\operatorname{sec}\theta$ C $-\operatorname{sec}\theta$ D $-\operatorname{csc}\theta$

Answer and solution

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

$270^\circ - \theta$ is of the form $90^\circ \pm \theta$ or $270^\circ \pm \theta$ , so $\sec$ changes to $\csc$

For acute $\theta$ , $270^\circ - \theta \in \text{III}_Q$ , where $\sec$ is -ve

$\therefore \sec(270^\circ - \theta) = -\csc\theta$
$\displaystyle \cot \frac{29\pi}{4} =$
A $\displaystyle \frac{1}{2}$ B $-1$ C $1$ D $\displaystyle \frac{\sqrt{3}}{2}$

Answer and solution

Answer: (C) $1$

$\displaystyle \frac{29\pi}{4} = 29 \times 45^\circ = 1305^\circ$

$\cot 1305^\circ = \cot(1080^\circ + 225^\circ) = \cot 225^\circ$

$225^\circ \in \text{III}_Q$ , where $\cot$ is +ve

$\cot 225^\circ = \cot(180^\circ + 45^\circ) = +\cot 45^\circ = 1$
The range of $4\cos x$ is
A $[0, 4]$ B $[-4, 0]$ C $[-1, 1]$ D $[-4, 4]$

Answer and solution

Answer: (D) $[-4, 4]$

$-1 \le \cos x \le 1 \Rightarrow -4 \le 4\cos x \le 4$
The graph of $\sec x$ repeats after every
A $2\pi$ B $\pi$ C $\displaystyle \frac{\pi}{2}$ D $\displaystyle \frac{3\pi}{2}$

Answer and solution

Answer: (A) $2\pi$

$\displaystyle \sec x = \frac{1}{\cos x}$ and $\cos x$ has period $2\pi$ , so the graph of $\sec x$ repeats after $2\pi$
If $ABCD$ is a cyclic quadrilateral, then $\sec B + \sec D =$
A $-1$ B $0$ C $1$ D $2\operatorname{sec} B$

Answer and solution

Answer: (B) $0$

$B + D = 180^\circ \Rightarrow \sec D = \sec(180^\circ - B) = -\sec B$

$\sec B + \sec D = 0$
If $\displaystyle \sin\theta\cos\theta = \frac{1}{6}$ , then $\sin^6\theta + \cos^6\theta =$
A $\displaystyle \frac{17}{18}$ B $\displaystyle \frac{1}{12}$ C $\displaystyle \frac{35}{36}$ D $\displaystyle \frac{11}{12}$

Answer and solution

Answer: (D) $\displaystyle \frac{11}{12}$

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

$\displaystyle = 1 - 3\left(\frac{1}{6}\right)^2 = 1 - \frac{1}{12} = \frac{11}{12}$
$\sqrt{1 - \cos^2 250^\circ} =$
A $-\operatorname{cos} 250^\circ$ B $-\operatorname{sin} 250^\circ$ C $\operatorname{sin} 250^\circ$ D $\operatorname{cos} 250^\circ$

Answer and solution

Answer: (B) $-\operatorname{sin} 250^\circ$

$= \sqrt{\sin^2 250^\circ} = |\sin 250^\circ|$

$250^\circ \in \text{III}_Q \Rightarrow \sin 250^\circ < 0$ , so $= -\sin 250^\circ$
$\sin 70^\circ\cos 25^\circ - \cos 70^\circ\sin 25^\circ =$
A $\operatorname{sin} 95^\circ$ B $\displaystyle \frac{1}{2}$ C $\displaystyle \frac{1}{\sqrt{2}}$ D $\displaystyle \frac{\sqrt{3}}{2}$

Answer and solution

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

$\displaystyle = \sin(70^\circ - 25^\circ) = \sin 45^\circ = \frac{1}{\sqrt{2}}$
If $\cot A = 4$ and $\cot B = 3$ , then $\cot(A + B) =$
A $\displaystyle \frac{13}{7}$ B $\displaystyle \frac{7}{11}$ C $-13$ D $\displaystyle \frac{11}{7}$

Answer and solution

Answer: (D) $\displaystyle \frac{11}{7}$

$\displaystyle \cot(A + B) = \frac{\cot A\cot B - 1}{\cot B + \cot A} = \frac{12 - 1}{3 + 4} = \frac{11}{7}$
$\cos(45^\circ + \theta)\cos(45^\circ - \theta) =$
A $\displaystyle \frac{1}{2} - \operatorname{cos}^2\theta$ B $\displaystyle \frac{1}{2} + \operatorname{sin}^2\theta$ C $\displaystyle \frac{1}{2} - \operatorname{sin}^2\theta$ D $1 - \operatorname{sin}^2\theta$

Answer and solution

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

$\displaystyle = \cos^2 45^\circ - \sin^2\theta = \frac{1}{2} - \sin^2\theta$
$\sin(A + B + C) =$
A $\Sigma\operatorname{cos} A\operatorname{sin} B\operatorname{sin} C - \Pi\operatorname{sin} A$ B $\Sigma\operatorname{sin} A\operatorname{cos} B\operatorname{cos} C - \Pi\operatorname{sin} A$ C $\Sigma\operatorname{sin} A\operatorname{cos} B\operatorname{cos} C + \Pi\operatorname{sin} A$ D $\Pi\operatorname{cos} A - \Sigma\operatorname{cos} A\operatorname{sin} B\operatorname{sin} C$

Answer and solution

Answer: (B) $\Sigma\operatorname{sin} A\operatorname{cos} B\operatorname{cos} C - \Pi\operatorname{sin} A$

$\sin[(A + B) + C] = \sin(A + B)\cos C + \cos(A + B)\sin C$

$= \sin A\cos B\cos C + \cos A\sin B\cos C + \cos A\cos B\sin C - \sin A\sin B\sin C$
$(1 + \cot 15^\circ)(1 + \cot 30^\circ) =$
A $3 + 2\sqrt{3}$ B $6 + 4\sqrt{3}$ C $2$ D $4$

Answer and solution

Answer: (B) $6 + 4\sqrt{3}$

$15^\circ + 30^\circ = 45^\circ \Rightarrow (1 + \cot A)(1 + \cot B) = 2\cot A\cot B$

$= 2(2 + \sqrt{3})\sqrt{3} = 6 + 4\sqrt{3}$
If $\displaystyle \tan A = \frac{1}{3}$ , then $\cos 2A =$
A $\displaystyle \frac{3}{5}$ B $\displaystyle \frac{8}{9}$ C $\displaystyle \frac{4}{5}$ D $\displaystyle \frac{3}{4}$

Answer and solution

Answer: (C) $\displaystyle \frac{4}{5}$

$\displaystyle \cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - \frac{1}{9}}{1 + \frac{1}{9}} = \frac{8}{10} = \frac{4}{5}$
If $\displaystyle \cos A = \frac{7}{25}$ and $A$ is acute, then $\displaystyle \tan\frac{A}{2} =$
A $\displaystyle \frac{4}{3}$ B $\displaystyle \frac{3}{4}$ C $\displaystyle \frac{9}{16}$ D $\displaystyle \frac{3}{5}$

Answer and solution

Answer: (B) $\displaystyle \frac{3}{4}$

$\displaystyle \tan\frac{A}{2} = \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \sqrt{\frac{18/25}{32/25}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$
$\sin 126^\circ =$
A $\displaystyle \frac{\sqrt{5}+1}{4}$ B $\displaystyle -\frac{\sqrt{5}+1}{4}$ C $\displaystyle \frac{\sqrt{5}-1}{4}$ D $\displaystyle \frac{\sqrt{10-2\sqrt{5}}}{4}$

Answer and solution

Answer: (A) $\displaystyle \frac{\sqrt{5}+1}{4}$

$\displaystyle \sin 126^\circ = \sin(180^\circ - 54^\circ) = \sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5}+1}{4}$
$\cot A - \tan A =$
A $2\operatorname{tan} 2A$ B $2\operatorname{csc} 2A$ C $2\operatorname{cot} 2A$ D $\operatorname{cot} 2A$

Answer and solution

Answer: (C) $2\operatorname{cot} 2A$

$\displaystyle \frac{\cos A}{\sin A} - \frac{\sin A}{\cos A} = \frac{\cos^2 A - \sin^2 A}{\sin A\cos A} = \frac{\cos 2A}{\frac{1}{2}\sin 2A} = 2\cot 2A$
$\cos\theta + \cos 3\theta + \cos 5\theta + \cos 7\theta =$
A $\displaystyle \frac{\operatorname{sin} 8\theta}{2\operatorname{sin}\theta}$ B $\displaystyle \frac{\operatorname{sin} 4\theta}{2\operatorname{sin}\theta}$ C $4\operatorname{cos} 4\theta$ D $\displaystyle \frac{\operatorname{sin} 8\theta}{\operatorname{sin}\theta}$

Answer and solution

Answer: (A) $\displaystyle \frac{\operatorname{sin} 8\theta}{2\operatorname{sin}\theta}$

$a_1 = \theta$ , $a_n = 7\theta$ , $d = 2\theta$ , $n = 4$

Sum $\displaystyle = \frac{\cos 4\theta\sin 4\theta}{\sin\theta} = \frac{\sin 8\theta}{2\sin\theta}$
$\cos^3 20^\circ + \cos^3 140^\circ + \cos^3 100^\circ =$
A $\displaystyle \frac{3}{4}$ B $0$ C $\displaystyle \frac{1}{8}$ D $\displaystyle \frac{3}{8}$

Answer and solution

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

$\displaystyle \cos^3\theta + \cos^3(120^\circ + \theta) + \cos^3(120^\circ - \theta) = \frac{3}{4}\cos 3\theta$ with $\theta = 20^\circ$

$\displaystyle = \frac{3}{4}\cos 60^\circ = \frac{3}{8}$
$\sqrt{1 - \sin 20^\circ} =$
A $\operatorname{cos} 10^\circ - \operatorname{sin} 10^\circ$ B $\operatorname{cos} 20^\circ - \operatorname{sin} 20^\circ$ C $\operatorname{cos} 10^\circ + \operatorname{sin} 10^\circ$ D $\operatorname{sin} 10^\circ - \operatorname{cos} 10^\circ$

Answer and solution

Answer: (A) $\operatorname{cos} 10^\circ - \operatorname{sin} 10^\circ$

$\sqrt{1 - \sin 2A} = |\cos A - \sin A|$ with $A = 10^\circ$

$\cos 10^\circ > \sin 10^\circ$ , so $= \cos 10^\circ - \sin 10^\circ$
$\displaystyle 2\cos 52\tfrac{1}{2}^\circ\cos 7\tfrac{1}{2}^\circ =$
A $\displaystyle \frac{\sqrt{3} + \sqrt{2}}{2}$ B $\displaystyle \frac{\sqrt{2} - 1}{2}$ C $\displaystyle \frac{1}{2}$ D $\displaystyle \frac{1 + \sqrt{2}}{2}$

Answer and solution

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

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

$\displaystyle = \cos 60^\circ + \cos 45^\circ = \frac{1}{2} + \frac{1}{\sqrt{2}} = \frac{1 + \sqrt{2}}{2}$
The period of $\tan 2x + \sin 4x$ is
A $2\pi$ B $\pi$ C $\displaystyle \frac{\pi}{4}$ D $\displaystyle \frac{\pi}{2}$

Answer and solution

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

Periods: $\displaystyle \tan 2x \to \frac{\pi}{2}$ , $\displaystyle \sin 4x \to \frac{\pi}{2}$ ; L.C.M $\displaystyle = \frac{\pi}{2}$

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MCQ test, Set 6