Advanced Trigonometry: MCQ test, Set 2

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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.

$(1 - \cos^2\theta)(1 + \cot^2\theta) =$
A $1$ B $\operatorname{sin}^2\theta$ C $0$ D $\operatorname{cos}^2\theta$

Answer and solution

Answer: (A) $1$

$1 - \cos^2\theta = \sin^2\theta$ and $1 + \cot^2\theta = \csc^2\theta$

$\sin^2\theta \cdot \csc^2\theta = 1$
$\displaystyle \frac{5\pi}{9}$ radians in degrees is
A $200^\circ$ B $120^\circ$ C $50^\circ$ D $100^\circ$

Answer and solution

Answer: (D) $100^\circ$

$\displaystyle \frac{5\pi}{9} = \frac{5}{9} \times 180^\circ = 100^\circ$
If $\operatorname{sin}\theta < 0$ and $\operatorname{cot}\theta > 0$ , then $\theta$ lies in
A $\text{IV}_Q$ B $\text{III}_Q$ C $\text{I}_Q$ D $\text{II}_Q$

Answer and solution

Answer: (B) $\text{III}_Q$

$\sin$ is -ve and $\cot$ is +ve only in $\text{III}_Q$
$\tan^2 60^\circ + 2\tan^2 45^\circ =$
A $5$ B $\displaystyle \frac{7}{3}$ C $4$ D $3$

Answer and solution

Answer: (A) $5$

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

$= 5$
$\operatorname{cos}(270^\circ - \theta) =$
A $-\operatorname{cos}\theta$ B $\operatorname{cos}\theta$ C $-\operatorname{sin}\theta$ D $\operatorname{sin}\theta$

Answer and solution

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

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

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

$\therefore \cos(270^\circ - \theta) = -\sin\theta$
$\tan 930^\circ =$
A $\displaystyle -\frac{1}{\sqrt{3}}$ B $\sqrt{3}$ C $\displaystyle \frac{1}{\sqrt{3}}$ D $-\sqrt{3}$

Answer and solution

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

$\tan 930^\circ = \tan(720^\circ + 210^\circ) = \tan 210^\circ$

$210^\circ \in \text{III}_Q$ , where $\tan$ is +ve

$\displaystyle \tan 210^\circ = \tan(180^\circ + 30^\circ) = +\tan 30^\circ = \frac{1}{\sqrt{3}}$
For which value of $k$ is $\sin\theta = k$ impossible?
A $\displaystyle \frac{4}{3}$ B $-1$ C $\displaystyle -\frac{1}{2}$ D $\displaystyle \frac{3}{4}$

Answer and solution

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

$-1 \le \sin\theta \le 1$

$\displaystyle \frac{4}{3} > 1$ , so $\displaystyle \sin\theta = \frac{4}{3}$ is impossible
From its graph, the period of $\cot x$ is
A $2\pi$ B $\pi$ C $\displaystyle \frac{\pi}{2}$ D $4\pi$

Answer and solution

Answer: (B) $\pi$

The graph of $\cot x$ repeats after every interval of length $\pi$ (asymptotes at $0, \pi, 2\pi$ )
$\sec 70^\circ - \csc 20^\circ =$
A $0$ B $-1$ C $2\operatorname{sec} 70^\circ$ D $1$

Answer and solution

Answer: (A) $0$

$70^\circ + 20^\circ = 90^\circ \Rightarrow \sec 70^\circ = \csc 20^\circ$

Difference $= 0$
If $\displaystyle \sin\theta\cos\theta = \frac{1}{5}$ , then $\sin^4\theta + \cos^4\theta =$
A $\displaystyle \frac{24}{25}$ B $\displaystyle \frac{23}{25}$ C $\displaystyle \frac{21}{25}$ D $\displaystyle \frac{22}{25}$

Answer and solution

Answer: (B) $\displaystyle \frac{23}{25}$

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

$\displaystyle = 1 - 2\left(\frac{1}{5}\right)^2 = 1 - \frac{2}{25} = \frac{23}{25}$
$\sqrt{1 - \sin^2 130^\circ} =$
A $\operatorname{sin} 130^\circ$ B $-\operatorname{sin} 130^\circ$ C $-\operatorname{cos} 130^\circ$ D $\operatorname{cos} 130^\circ$

Answer and solution

Answer: (C) $-\operatorname{cos} 130^\circ$

$= \sqrt{\cos^2 130^\circ} = |\cos 130^\circ|$

$130^\circ \in \text{II}_Q \Rightarrow \cos 130^\circ < 0$ , so $= -\cos 130^\circ$
$\sin 40^\circ\cos 20^\circ + \cos 40^\circ\sin 20^\circ =$
A $\operatorname{sin} 20^\circ$ B $\displaystyle \frac{\sqrt{3}}{2}$ C $1$ D $\displaystyle \frac{1}{2}$

Answer and solution

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

$\displaystyle = \sin(40^\circ + 20^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$
$\displaystyle \frac{1 + \tan 15^\circ}{1 - \tan 15^\circ} =$
A $2 + \sqrt{3}$ B $1$ C $\displaystyle \frac{1}{\sqrt{3}}$ D $\sqrt{3}$

Answer and solution

Answer: (D) $\sqrt{3}$

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

$= \tan 60^\circ = \sqrt{3}$
$\sin^2 105^\circ - \sin^2 15^\circ =$
A $\displaystyle \frac{1}{2}$ B $\displaystyle -\frac{\sqrt{3}}{2}$ C $\displaystyle \frac{\sqrt{3}}{2}$ D $1$

Answer and solution

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

$\displaystyle = \sin 120^\circ \sin 90^\circ = \frac{\sqrt{3}}{2}$
$\tan 20^\circ + \tan 40^\circ + \tan 120^\circ =$
A $0$ B $-\sqrt{3}\operatorname{tan} 20^\circ\operatorname{tan} 40^\circ$ C $\operatorname{tan} 20^\circ\operatorname{tan} 40^\circ$ D $\sqrt{3}\operatorname{tan} 20^\circ\operatorname{tan} 40^\circ$

Answer and solution

Answer: (B) $-\sqrt{3}\operatorname{tan} 20^\circ\operatorname{tan} 40^\circ$

$20^\circ + 40^\circ + 120^\circ = 180^\circ$ , so $\Sigma\tan A = \Pi\tan A$

$= \tan 20^\circ\tan 40^\circ\tan 120^\circ = -\sqrt{3}\tan 20^\circ\tan 40^\circ$
$(1 + \tan 1^\circ)(1 + \tan 2^\circ)(1 + \tan 3^\circ)\cdots(1 + \tan 44^\circ) =$
A $2^{11}$ B $2^{44}$ C $44$ D $2^{22}$

Answer and solution

Answer: (D) $2^{22}$

Pair $(1^\circ, 44^\circ), (2^\circ, 43^\circ), \ldots, (22^\circ, 23^\circ)$ : each pair gives $2$

22 pairs $\Rightarrow 2^{22}$
$\displaystyle \frac{1 - \cos 2A}{1 + \cos 2A} =$
A $\operatorname{tan}^2 A$ B $\operatorname{cot}^2 A$ C $\operatorname{tan} A$ D $\operatorname{sin}^2 A$

Answer and solution

Answer: (A) $\operatorname{tan}^2 A$

$1 - \cos 2A = 2\sin^2 A$ , $1 + \cos 2A = 2\cos^2 A$

Ratio $= \tan^2 A$
If $\displaystyle \cos A = \frac{1}{3}$ , then $\cos 3A =$
A $\displaystyle -\frac{23}{27}$ B $\displaystyle -\frac{5}{9}$ C $\displaystyle \frac{4}{27}$ D $\displaystyle \frac{23}{27}$

Answer and solution

Answer: (A) $\displaystyle -\frac{23}{27}$

$\displaystyle \cos 3A = 4\cos^3 A - 3\cos A = \frac{4}{27} - 1 = -\frac{23}{27}$
$\displaystyle \tan 22\tfrac{1}{2}^\circ =$
A $\displaystyle \frac{1}{\sqrt{2}}$ B $\sqrt{2} - 1$ C $2 - \sqrt{3}$ D $\sqrt{2} + 1$

Answer and solution

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

$\displaystyle \tan^2 22\tfrac{1}{2}^\circ = \frac{1 - \cos 45^\circ}{1 + \cos 45^\circ} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = (\sqrt{2} - 1)^2$

$\displaystyle \therefore \tan 22\tfrac{1}{2}^\circ = \sqrt{2} - 1$
$\displaystyle \cos\frac{\pi}{5}\cos\frac{2\pi}{5} =$
A $\displaystyle \frac{1}{8}$ B $\displaystyle \frac{1}{2}$ C $\displaystyle \frac{1}{4}$ D $\displaystyle -\frac{1}{4}$

Answer and solution

Answer: (C) $\displaystyle \frac{1}{4}$

$\displaystyle \cos\theta\cos 2\theta = \frac{\sin 4\theta}{4\sin\theta}$

$\displaystyle = \frac{\sin\frac{4\pi}{5}}{4\sin\frac{\pi}{5}} = \frac{1}{4}$
$\displaystyle \sin\frac{\pi}{5} + \sin\frac{2\pi}{5} + \sin\frac{3\pi}{5} + \sin\frac{4\pi}{5} =$
A $\operatorname{cot} 18^\circ$ B $2\operatorname{cot} 18^\circ$ C $\operatorname{tan} 18^\circ$ D $1$

Answer and solution

Answer: (A) $\operatorname{cot} 18^\circ$

$\displaystyle \frac{\sin\frac{a_1 + a_n}{2}\sin\frac{nd}{2}}{\sin\frac{d}{2}} = \frac{\sin\frac{\pi}{2}\sin\frac{2\pi}{5}}{\sin\frac{\pi}{10}}$

$\displaystyle = \frac{\sin 72^\circ}{\sin 18^\circ} = \frac{\cos 18^\circ}{\sin 18^\circ} = \cot 18^\circ$
$\tan 6^\circ\tan 54^\circ\tan 66^\circ =$
A $\operatorname{tan} 6^\circ$ B $\operatorname{cot} 18^\circ$ C $3\operatorname{tan} 18^\circ$ D $\operatorname{tan} 18^\circ$

Answer and solution

Answer: (D) $\operatorname{tan} 18^\circ$

$\tan\theta\tan(60^\circ - \theta)\tan(60^\circ + \theta) = \tan 3\theta$ with $\theta = 6^\circ$

$= \tan 18^\circ$
If $45^\circ < A < 90^\circ$ , then $\sqrt{1 - \sin 2A} =$
A $\operatorname{cos} 2A$ B $\operatorname{sin} A + \operatorname{cos} A$ C $\operatorname{sin} A - \operatorname{cos} A$ D $\operatorname{cos} A - \operatorname{sin} A$

Answer and solution

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

$\sqrt{1 - \sin 2A} = |\cos A - \sin A|$

For $45^\circ < A < 90^\circ$ , $\sin A > \cos A$ , so $= \sin A - \cos A$
$\cos 75^\circ + \cos 15^\circ =$
A $\displaystyle \frac{\sqrt{3}}{2}$ B $\sqrt{2}$ C $\displaystyle \frac{1}{\sqrt{2}}$ D $\displaystyle \frac{\sqrt{3}}{\sqrt{2}}$

Answer and solution

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

$\displaystyle \cos C + \cos D = 2\cos\frac{C + D}{2}\cos\frac{C - D}{2}$

$\displaystyle = 2\cos 45^\circ\cos 30^\circ = 2\cdot\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{\sqrt{2}}$
The period of $\cot(7 - 2x)$ is
A $\displaystyle \frac{\pi}{7}$ B $2\pi$ C $\pi$ D $\displaystyle \frac{\pi}{2}$

Answer and solution

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

Period of $\cot(ax + b)$ is $\displaystyle \frac{\pi}{|a|} = \frac{\pi}{|-2|} = \frac{\pi}{2}$

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MCQ test, Set 2