Advanced Trigonometry: MCQ test, Set 5

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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.

If $\displaystyle \cos\theta = \frac{7}{25}$ and $\theta$ is acute, then $\csc\theta =$
A $\displaystyle \frac{24}{7}$ B $\displaystyle \frac{25}{24}$ C $\displaystyle \frac{24}{25}$ D $\displaystyle \frac{25}{7}$

Answer and solution

Answer: (B) $\displaystyle \frac{25}{24}$

$\displaystyle \cos\theta = \frac{\text{adj}}{\text{hyp}} = \frac{7}{25}$ : take adj $= 7$ , hyp $= 25$

opp $= \sqrt{25^2 - 7^2} = \sqrt{576} = 24$

$\displaystyle \csc\theta = \frac{\text{hyp}}{\text{opp}} = \frac{25}{24}$
Four complete rotations measure
A $16\pi \text{ radians}$ B $4\pi \text{ radians}$ C $2\pi \text{ radians}$ D $8\pi \text{ radians}$

Answer and solution

Answer: (D) $8\pi \text{ radians}$

One complete rotation $= 360^\circ = 2\pi$

Four rotations $= 4 \times 2\pi = 8\pi$
If $\operatorname{cos}\theta < 0$ and $\operatorname{csc}\theta > 0$ , then $\theta$ lies in
A $\text{I}_Q$ B $\text{II}_Q$ C $\text{IV}_Q$ D $\text{III}_Q$

Answer and solution

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

$\cos$ is -ve and $\csc$ is +ve only in $\text{II}_Q$
$\displaystyle \frac{1 - \cos 60^\circ}{1 + \cos 60^\circ} =$
A $\displaystyle \frac{1}{3}$ B $\displaystyle \frac{1}{2}$ C $3$ D $2 - \sqrt{3}$

Answer and solution

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

$\displaystyle = \frac{1 - \frac{1}{2}}{1 + \frac{1}{2}} = \frac{1/2}{3/2}$

$\displaystyle = \frac{1}{3}$
$\operatorname{tan}(180^\circ + \theta) =$
A $\operatorname{cot}\theta$ B $-\operatorname{tan}\theta$ C $-\operatorname{cot}\theta$ D $\operatorname{tan}\theta$

Answer and solution

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

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

For acute $\theta$ , $180^\circ + \theta \in \text{III}_Q$ , where $\tan$ is +ve

$\therefore \tan(180^\circ + \theta) = \tan\theta$
$\sin 1590^\circ =$
A $\displaystyle -\frac{\sqrt{3}}{2}$ B $\displaystyle \frac{\sqrt{3}}{2}$ C $\displaystyle \frac{1}{2}$ D $\displaystyle -\frac{1}{2}$

Answer and solution

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

$\sin 1590^\circ = \sin(1440^\circ + 150^\circ) = \sin 150^\circ$

$150^\circ \in \text{II}_Q$ , where $\sin$ is +ve

$\displaystyle \sin 150^\circ = \sin(180^\circ - 30^\circ) = +\sin 30^\circ = \frac{1}{2}$
The least value of $5 + 2\sin x$ is
A $5$ B $7$ C $1$ D $3$

Answer and solution

Answer: (D) $3$

$-1 \le \sin x \le 1 \Rightarrow -2 \le 2\sin x \le 2$

$\Rightarrow 3 \le 5 + 2\sin x \le 7$
Arrange $\tan 4$ , $\tan 5$ , $\tan 6$ (angles in radians):
A $\operatorname{tan} 4 > \operatorname{tan} 5 > \operatorname{tan} 6$ B $\operatorname{tan} 4 > \operatorname{tan} 6 > \operatorname{tan} 5$ C $\operatorname{tan} 5 > \operatorname{tan} 4 > \operatorname{tan} 6$ D $\operatorname{tan} 6 > \operatorname{tan} 5 > \operatorname{tan} 4$

Answer and solution

Answer: (B) $\operatorname{tan} 4 > \operatorname{tan} 6 > \operatorname{tan} 5$

$4 \approx 229^\circ$ , $5 \approx 286^\circ$ , $6 \approx 343^\circ$

$\tan 229^\circ = \tan 49^\circ > 0$ , $\tan 286^\circ = -\tan 74^\circ$ , $\tan 343^\circ = -\tan 17^\circ$

$-\tan 17^\circ > -\tan 74^\circ$ , so $\tan 4 > \tan 6 > \tan 5$
If $\tan 2A = \cot(A - 18^\circ)$ , where $2A$ is acute, then $A =$
A $18^\circ$ B $24^\circ$ C $36^\circ$ D $72^\circ$

Answer and solution

Answer: (C) $36^\circ$

$\tan 2A = \cot(A - 18^\circ) \Rightarrow 2A + (A - 18^\circ) = 90^\circ$

$3A = 108^\circ \Rightarrow A = 36^\circ$
The value of $\cos^2\theta + \sin^4\theta$ at $\theta = 60^\circ$ is
A $\displaystyle \frac{13}{16}$ B $\displaystyle \frac{5}{8}$ C $1$ D $\displaystyle \frac{7}{16}$

Answer and solution

Answer: (A) $\displaystyle \frac{13}{16}$

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

$\displaystyle = 1 - \frac{3}{4}\cdot\frac{1}{4} = \frac{13}{16}$
The least value of $\tan^2\theta + 49\cot^2\theta$ is
A $2$ B $14$ C $7$ D $50$

Answer and solution

Answer: (B) $14$

Consider $\tan^2\theta$ and $49\cot^2\theta$ ; both are +ve

A.M $\ge$ G.M: $\displaystyle \frac{\tan^2\theta + 49\cot^2\theta}{2} \ge \sqrt{\tan^2\theta \cdot 49\cot^2\theta} = \sqrt{49} = 7$

$\Rightarrow \tan^2\theta + 49\cot^2\theta \ge 14$ ; equality when $\tan^2\theta = 7$
$\sin 165^\circ =$
A $\displaystyle \frac{\sqrt{3}-1}{2\sqrt{2}}$ B $\displaystyle \frac{\sqrt{3}+1}{2\sqrt{2}}$ C $\displaystyle \frac{1-\sqrt{3}}{2\sqrt{2}}$ D $\displaystyle \frac{\sqrt{3}-1}{2}$

Answer and solution

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

$\sin 165^\circ = \sin(180^\circ - 15^\circ) = \sin 15^\circ$

$\displaystyle \sin(45^\circ - 30^\circ) = \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\cdot\frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$
If $\displaystyle \tan A = \frac{1}{2}$ and $\displaystyle \tan B = \frac{1}{3}$ ( $A$ , $B$ acute), then $A + B =$
A $\displaystyle \frac{\pi}{4}$ B $\displaystyle \frac{\pi}{6}$ C $\displaystyle \frac{\pi}{2}$ D $\displaystyle \frac{\pi}{3}$

Answer and solution

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

$\displaystyle \tan(A + B) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{6}} = \frac{5/6}{5/6} = 1$

$\displaystyle \therefore A + B = \frac{\pi}{4}$
$\sin(60^\circ + \theta)\sin(60^\circ - \theta) =$
A $\displaystyle \frac{1}{4} - \operatorname{sin}^2\theta$ B $\displaystyle \frac{3}{4} - \operatorname{sin}^2\theta$ C $\displaystyle \frac{3}{4} - \operatorname{cos}^2\theta$ D $\displaystyle \frac{3}{4} + \operatorname{sin}^2\theta$

Answer and solution

Answer: (B) $\displaystyle \frac{3}{4} - \operatorname{sin}^2\theta$

$\displaystyle = \sin^2 60^\circ - \sin^2\theta = \frac{3}{4} - \sin^2\theta$
In $\triangle ABC$ , $\cot A\cot B + \cot B\cot C + \cot C\cot A =$
A $0$ B $\operatorname{cot} A\operatorname{cot} B\operatorname{cot} C$ C $1$ D $-1$

Answer and solution

Answer: (C) $1$

$\cot(A + B) = \cot(180^\circ - C) = -\cot C$

$\displaystyle \frac{\cot A\cot B - 1}{\cot B + \cot A} = -\cot C \Rightarrow \Sigma\cot A\cot B = 1$
$(1 + \tan 11^\circ)(1 + \tan 34^\circ)(1 + \tan 14^\circ)(1 + \tan 31^\circ) =$
A $8$ B $2$ C $16$ D $4$

Answer and solution

Answer: (D) $4$

$11^\circ + 34^\circ = 45^\circ$ and $14^\circ + 31^\circ = 45^\circ$

$= 2 \times 2 = 4$
$1 - 2\sin^2 75^\circ =$
A $\displaystyle -\frac{\sqrt{3}}{2}$ B $\displaystyle \frac{1}{2}$ C $\displaystyle -\frac{1}{2}$ D $\displaystyle \frac{\sqrt{3}}{2}$

Answer and solution

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

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

$\displaystyle = \cos 150^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$
$4\cos^3 40^\circ - 3\cos 40^\circ =$
A $\displaystyle -\frac{1}{2}$ B $\operatorname{cos} 40^\circ$ C $\displaystyle \frac{1}{2}$ D $\displaystyle -\frac{\sqrt{3}}{2}$

Answer and solution

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

$4\cos^3 A - 3\cos A = \cos 3A$

$\displaystyle = \cos 120^\circ = -\frac{1}{2}$
$\displaystyle \tan 22\tfrac{1}{2}^\circ + \cot 22\tfrac{1}{2}^\circ =$
A $2\sqrt{3}$ B $2\sqrt{2}$ C $\sqrt{2}$ D $2$

Answer and solution

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

$\displaystyle \tan 22\tfrac{1}{2}^\circ = \sqrt{2} - 1$ , $\displaystyle \cot 22\tfrac{1}{2}^\circ = \sqrt{2} + 1$

Sum $= 2\sqrt{2}$
$\displaystyle \cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7} =$
A $\displaystyle \frac{1}{8}$ B $\displaystyle -\frac{1}{8}$ C $\displaystyle \frac{1}{16}$ D $\displaystyle -\frac{1}{4}$

Answer and solution

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

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

$\displaystyle \sin\frac{8\pi}{7} = -\sin\frac{\pi}{7}$ , so G.E $\displaystyle = -\frac{1}{8}$
$\sin x + \sin 3x + \sin 5x =$
A $\displaystyle \frac{\operatorname{sin} 9x}{\operatorname{sin} x}$ B $\displaystyle \frac{\operatorname{sin} 6x}{2\operatorname{sin} x}$ C $\displaystyle \frac{\operatorname{sin}^2 3x}{\operatorname{sin} x}$ D $3\operatorname{sin} 3x$

Answer and solution

Answer: (C) $\displaystyle \frac{\operatorname{sin}^2 3x}{\operatorname{sin} x}$

$a_1 = x$ , $a_n = 5x$ , $d = 2x$ , $n = 3$

Sum $\displaystyle = \frac{\sin 3x\sin 3x}{\sin x} = \frac{\sin^2 3x}{\sin x}$
$\sin^2 12^\circ + \sin^2 48^\circ + \sin 12^\circ\sin 48^\circ =$
A $\displaystyle \frac{3}{2}$ B $1$ C $\displaystyle \frac{1}{2}$ D $\displaystyle \frac{3}{4}$

Answer and solution

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

$12^\circ + 48^\circ = 60^\circ$

If $A + B = 60^\circ$ , $\displaystyle \sin^2 A + \sin^2 B + \sin A\sin B = \frac{3}{4}$
If $\displaystyle \frac{x}{y} = \frac{\tan 67\tfrac{1}{2}^\circ}{\tan 22\tfrac{1}{2}^\circ}$ , then $\displaystyle \frac{x + y}{x - y} =$
A $\sqrt{2} + 1$ B $\displaystyle \frac{1}{\sqrt{2}}$ C $\sqrt{2}$ D $2$

Answer and solution

Answer: (C) $\sqrt{2}$

$\displaystyle \frac{x + y}{x - y} = \frac{\sin(A + B)}{\sin(A - B)} = \frac{\sin 90^\circ}{\sin 45^\circ} = \sqrt{2}$
$\sin 75^\circ - \sin 15^\circ =$
A $\displaystyle \frac{\sqrt{3}}{\sqrt{2}}$ B $0$ C $\displaystyle \frac{1}{\sqrt{2}}$ D $\displaystyle \frac{1}{2}$

Answer and solution

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

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

$\displaystyle = 2\cos 45^\circ\sin 30^\circ = \frac{1}{\sqrt{2}}$
The period of $\displaystyle \sin\frac{x}{4} + \cos\frac{x}{6}$ is
A $10\pi$ B $12\pi$ C $48\pi$ D $24\pi$

Answer and solution

Answer: (D) $24\pi$

Periods $8\pi$ and $12\pi$ ; L.C.M $= 24\pi$

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MCQ test, Set 5