Advanced Trigonometry: MCQ test, Set 4

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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.

$(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) =$
A $-1$ B $0$ C $1$ D $2\operatorname{sec}\theta$

Answer and solution

Answer: (C) $1$

$= \sec^2\theta - \tan^2\theta = 1$
$180^g$ in degrees is
A $172^\circ$ B $200^\circ$ C $162^\circ$ D $180^\circ$

Answer and solution

Answer: (C) $162^\circ$

$\displaystyle 1^g = \frac{9}{10}$ degree

$\displaystyle 180^g = 180 \times \frac{9}{10} = 162^\circ$
If $\theta = 310^\circ$ , which of the following is positive?
A $\operatorname{sin}\theta$ B $\operatorname{cot}\theta$ C $\operatorname{tan}\theta$ D $\operatorname{cos}\theta$

Answer and solution

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

$310^\circ \in \text{IV}_Q$ , where only cos, sec are +ve
$\cot 30^\circ \cdot \tan 60^\circ =$
A $\displaystyle \frac{1}{3}$ B $1$ C $3$ D $\sqrt{3}$

Answer and solution

Answer: (C) $3$

$= \sqrt{3} \times \sqrt{3}$

$= 3$
$\operatorname{sin}(270^\circ + \theta) =$
A $-\operatorname{cos}\theta$ B $\operatorname{cos}\theta$ C $\operatorname{sin}\theta$ D $-\operatorname{sin}\theta$

Answer and solution

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

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

For acute $\theta$ , $270^\circ + \theta \in \text{IV}_Q$ , where $\sin$ is -ve

$\therefore \sin(270^\circ + \theta) = -\cos\theta$
$\csc 1035^\circ =$
A $-\sqrt{2}$ B $\displaystyle -\frac{1}{\sqrt{2}}$ C $\displaystyle \frac{1}{\sqrt{2}}$ D $\sqrt{2}$

Answer and solution

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

$\csc 1035^\circ = \csc(720^\circ + 315^\circ) = \csc 315^\circ$

$315^\circ \in \text{IV}_Q$ , where $\csc$ is -ve

$\csc 315^\circ = \csc(360^\circ - 45^\circ) = -\csc 45^\circ = -\sqrt{2}$
$\csc\theta$ can never be equal to
A $2$ B $-1$ C $\displaystyle -\frac{3}{2}$ D $\displaystyle \frac{1}{2}$

Answer and solution

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

$\csc\theta \le -1$ or $\csc\theta \ge 1$

$\displaystyle \frac{1}{2}$ lies between $-1$ and $1$ , so it is never a value of $\csc\theta$
Between $0$ and $2\pi$ , the graph of $\sin x$ reaches its least value at $x =$
A $\displaystyle \frac{3\pi}{2}$ B $\pi$ C $2\pi$ D $\displaystyle \frac{\pi}{2}$

Answer and solution

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

The least value of $\sin x$ is $-1$

$\displaystyle \sin\frac{3\pi}{2} = \sin 270^\circ = -1$
$\cos^2 20^\circ + \cos^2 70^\circ =$
A $2$ B $0$ C $2\operatorname{cos}^2 20^\circ$ D $1$

Answer and solution

Answer: (D) $1$

$\cos 70^\circ = \sin 20^\circ$

$\cos^2 20^\circ + \sin^2 20^\circ = 1$
The value of $\sin^4\theta + \cos^4\theta$ at $\theta = 45^\circ$ is
A $1$ B $\displaystyle \frac{1}{2}$ C $\displaystyle \frac{1}{4}$ D $0$

Answer and solution

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

$\displaystyle = 1 - 2\sin^2 45^\circ\cos^2 45^\circ = 1 - 2\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{2}$
If $x < 0$ , the greatest value of $\displaystyle x + \frac{1}{x}$ is
A $-1$ B $2$ C $0$ D $-2$

Answer and solution

Answer: (D) $-2$

For $x < 0$ , $-x$ and $\displaystyle -\frac{1}{x}$ are +ve

A.M $\ge$ G.M: $\displaystyle -x - \frac{1}{x} \ge 2 \Rightarrow x + \frac{1}{x} \le -2$
$\cos 25^\circ\cos 35^\circ - \sin 25^\circ\sin 35^\circ =$
A $\displaystyle \frac{\sqrt{3}}{2}$ B $\displaystyle \frac{1}{2}$ C $\operatorname{cos} 10^\circ$ D $0$

Answer and solution

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

$\displaystyle = \cos(25^\circ + 35^\circ) = \cos 60^\circ = \frac{1}{2}$
$\displaystyle \frac{\tan 65^\circ - \tan 20^\circ}{1 + \tan 65^\circ\tan 20^\circ} =$
A $0$ B $1$ C $\sqrt{3}$ D $\operatorname{tan} 85^\circ$

Answer and solution

Answer: (B) $1$

$= \tan(65^\circ - 20^\circ) = \tan 45^\circ = 1$
$\sin^2 24^\circ - \sin^2 6^\circ =$
A $\displaystyle \frac{\sqrt{5}-1}{8}$ B $\displaystyle \frac{1}{4}$ C $\displaystyle \frac{\sqrt{5}-1}{4}$ D $\displaystyle \frac{\sqrt{5}+1}{8}$

Answer and solution

Answer: (A) $\displaystyle \frac{\sqrt{5}-1}{8}$

$\displaystyle = \sin 30^\circ \sin 18^\circ = \frac{1}{2} \cdot \frac{\sqrt{5}-1}{4}$
In $\triangle ABC$ , if $\tan A = 1$ and $\tan B = 3$ , then $\tan C =$
A $2$ B $-2$ C $\displaystyle \frac{1}{2}$ D $4$

Answer and solution

Answer: (A) $2$

$1 + 3 + \tan C = 3\tan C$

$2\tan C = 4 \Rightarrow \tan C = 2$
$\tan 3^\circ + \tan 42^\circ + \tan 3^\circ\tan 42^\circ =$
A $-1$ B $2$ C $1$ D $0$

Answer and solution

Answer: (C) $1$

$3^\circ + 42^\circ = 45^\circ$ : $\tan A + \tan B = 1 - \tan A\tan B$

So the sum is $1$
If $\tan A = 2$ , then $\sin 2A =$
A $\displaystyle -\frac{4}{3}$ B $4$ C $\displaystyle \frac{4}{5}$ D $\displaystyle -\frac{3}{5}$

Answer and solution

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

$\displaystyle \sin 2A = \frac{2\tan A}{1 + \tan^2 A} = \frac{4}{1 + 4} = \frac{4}{5}$
If $\displaystyle \tan A = \frac{1}{2}$ , then $\tan 3A =$
A $\displaystyle \frac{11}{8}$ B $\displaystyle \frac{11}{2}$ C $\displaystyle \frac{13}{2}$ D $\displaystyle \frac{4}{3}$

Answer and solution

Answer: (B) $\displaystyle \frac{11}{2}$

$\displaystyle \tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A} = \frac{\frac{3}{2} - \frac{1}{8}}{1 - \frac{3}{4}} = \frac{11/8}{1/4} = \frac{11}{2}$
$\cos^2 36^\circ + \sin^2 18^\circ =$
A $\displaystyle \frac{3}{4}$ B $\displaystyle \frac{\sqrt{5}}{4}$ C $1$ D $\displaystyle \frac{1}{2}$

Answer and solution

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

$\displaystyle = \frac{(\sqrt{5}+1)^2 + (\sqrt{5}-1)^2}{16} = \frac{(6 + 2\sqrt{5}) + (6 - 2\sqrt{5})}{16} = \frac{12}{16} = \frac{3}{4}$
$\cot 75^\circ + \tan 75^\circ =$
A $2\sqrt{3}$ B $1$ C $2$ D $4$

Answer and solution

Answer: (D) $4$

$\cot A + \tan A = 2\csc 2A$

$= 2\csc 150^\circ = 2 \times 2 = 4$
$\displaystyle \sin\frac{2\pi}{7}\sin\frac{4\pi}{7}\sin\frac{6\pi}{7} =$
A $\displaystyle \frac{1}{8}$ B $\displaystyle \frac{\sqrt{7}}{8}$ C $\displaystyle \frac{\sqrt{7}}{4}$ D $\displaystyle \frac{7}{64}$

Answer and solution

Answer: (B) $\displaystyle \frac{\sqrt{7}}{8}$

$\displaystyle \sin\frac{4\pi}{7} = \sin\frac{3\pi}{7}$ , $\displaystyle \sin\frac{6\pi}{7} = \sin\frac{\pi}{7}$

G.E $\displaystyle = \sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7} = \frac{\sqrt{7}}{8}$
$\cos^2 50^\circ + \cos^2 10^\circ - \cos 50^\circ\cos 10^\circ =$
A $1$ B $\displaystyle \frac{3}{2}$ C $\displaystyle \frac{1}{4}$ D $\displaystyle \frac{3}{4}$

Answer and solution

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

$50^\circ + 10^\circ = 60^\circ$

If $A + B = 60^\circ$ , $\displaystyle \cos^2 A + \cos^2 B - \cos A\cos B = \frac{3}{4}$
$\sqrt{1 + \sin 70^\circ} + \sqrt{1 - \sin 70^\circ} =$
A $2\operatorname{sin} 35^\circ$ B $2\operatorname{cos} 35^\circ$ C $\operatorname{cos} 35^\circ$ D $2\operatorname{cos} 70^\circ$

Answer and solution

Answer: (B) $2\operatorname{cos} 35^\circ$

$= (\cos 35^\circ + \sin 35^\circ) + (\cos 35^\circ - \sin 35^\circ) = 2\cos 35^\circ$
$\cos 52^\circ + \cos 68^\circ + \cos 172^\circ =$
A $\displaystyle \frac{1}{2}$ B $1$ C $0$ D $2\operatorname{cos} 8^\circ$

Answer and solution

Answer: (C) $0$

$\cos 68^\circ + \cos 52^\circ = 2\cos 60^\circ\cos 8^\circ = \cos 8^\circ$

$\cos 172^\circ = \cos(180^\circ - 8^\circ) = -\cos 8^\circ$

Sum $= 0$
The period of $|\sec x|$ is
A $\pi$ B $4\pi$ C $2\pi$ D $\displaystyle \frac{\pi}{2}$

Answer and solution

Answer: (A) $\pi$

$|\sec(x + \pi)| = |-\sec x| = |\sec x|$ , so period $= \pi$

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MCQ test, Set 4