Advanced Trigonometry: MCQ test, Set 7

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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 $\displaystyle \tan\theta = \frac{21}{20}$ and $\theta$ is acute, then $\cos\theta =$
A $\displaystyle \frac{29}{21}$ B $\displaystyle \frac{29}{20}$ C $\displaystyle \frac{21}{29}$ D $\displaystyle \frac{20}{29}$

Answer and solution

Answer: (D) $\displaystyle \frac{20}{29}$

$\displaystyle \tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{21}{20}$ : take opp $= 21$ , adj $= 20$

hyp $= \sqrt{21^2 + 20^2} = \sqrt{841} = 29$

$\displaystyle \cos\theta = \frac{\text{adj}}{\text{hyp}} = \frac{20}{29}$
$1$ radian is approximately
A $60^\circ$ B $90^\circ$ C $45^\circ$ D $57^\circ$

Answer and solution

Answer: (D) $57^\circ$

$\displaystyle 1^c = \frac{180^\circ}{\pi} \approx 57^\circ$
If $\theta = 130^\circ$ , which of the following is positive?
A $\operatorname{sin}\theta$ B $\operatorname{tan}\theta$ C $\operatorname{cot}\theta$ D $\operatorname{cos}\theta$

Answer and solution

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

$130^\circ \in \text{II}_Q$ , where only sin, csc are +ve
$\sec^2 30^\circ =$
A $4$ B $\displaystyle \frac{2}{\sqrt{3}}$ C $\displaystyle \frac{4}{3}$ D $\displaystyle \frac{3}{4}$

Answer and solution

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

$\displaystyle \sec 30^\circ = \frac{2}{\sqrt{3}}$ , so $\displaystyle \sec^2 30^\circ = \frac{4}{3}$

$\displaystyle = \frac{4}{3}$
$\operatorname{csc}(180^\circ - \theta) =$
A $\operatorname{csc}\theta$ B $-\operatorname{csc}\theta$ C $\operatorname{sec}\theta$ D $-\operatorname{sec}\theta$

Answer and solution

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

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

For acute $\theta$ , $180^\circ - \theta \in \text{II}_Q$ , where $\csc$ is +ve

$\therefore \csc(180^\circ - \theta) = \csc\theta$
$\cos 1080^\circ - \sin 990^\circ =$
A $2$ B $0$ C $-2$ D $1$

Answer and solution

Answer: (A) $2$

$\cos 1080^\circ = \cos(3 \times 360^\circ) = 1$

$\sin 990^\circ = \sin(720^\circ + 270^\circ) = \sin 270^\circ = -1$

$1 - (-1)$

$= 2$
Which value can $\sec\theta$ never take?
A $\displaystyle \frac{5}{4}$ B $\displaystyle -\frac{3}{2}$ C $\displaystyle -\frac{2}{3}$ D $1$

Answer and solution

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

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

$\displaystyle -1 < -\frac{2}{3} < 1$ , so $\sec\theta$ cannot be $\displaystyle -\frac{2}{3}$
Which is true for $\sin 4$ , $\sin 5$ , $\sin 6$ (angles in radians)?
A $\operatorname{sin} 6 > \operatorname{sin} 5 > \operatorname{sin} 4$ B $\operatorname{sin} 4 > \operatorname{sin} 5 > \operatorname{sin} 6$ C $\operatorname{sin} 5 > \operatorname{sin} 4 > \operatorname{sin} 6$ D $\operatorname{sin} 6 > \operatorname{sin} 4 > \operatorname{sin} 5$

Answer and solution

Answer: (D) $\operatorname{sin} 6 > \operatorname{sin} 4 > \operatorname{sin} 5$

$\sin 229^\circ = -\sin 49^\circ$ , $\sin 286^\circ = -\sin 74^\circ$ , $\sin 343^\circ = -\sin 17^\circ$

$-\sin 17^\circ > -\sin 49^\circ > -\sin 74^\circ$
$\tan 25^\circ + \tan 155^\circ =$
A $1$ B $2\operatorname{tan} 25^\circ$ C $-2\operatorname{tan} 25^\circ$ D $0$

Answer and solution

Answer: (D) $0$

$\tan 155^\circ = \tan(180^\circ - 25^\circ) = -\tan 25^\circ$

Sum $= 0$
$2(\sin^6\theta + \cos^6\theta) - 3(\sin^4\theta + \cos^4\theta) =$
A $-5$ B $-1$ C $1$ D $0$

Answer and solution

Answer: (B) $-1$

Let $k = \sin^2\theta\cos^2\theta$ : $2(1 - 3k) - 3(1 - 2k)$

$= 2 - 6k - 3 + 6k = -1$
The least value of $2\tan^2\theta + 8\cot^2\theta$ is
A $10$ B $8$ C $16$ D $4$

Answer and solution

Answer: (B) $8$

Consider $2\tan^2\theta$ and $8\cot^2\theta$ ; both are +ve

A.M $\ge$ G.M: $\displaystyle \frac{2\tan^2\theta + 8\cot^2\theta}{2} \ge \sqrt{2\tan^2\theta \cdot 8\cot^2\theta} = \sqrt{16} = 4$

$\Rightarrow 2\tan^2\theta + 8\cot^2\theta \ge 8$ ; equality when $\tan^2\theta = 2$
If $\displaystyle \sin A = \frac{4}{5}$ and $\displaystyle \cos B = \frac{15}{17}$ ( $A$ , $B$ acute), then $\cos(A - B) =$
A $\displaystyle \frac{77}{85}$ B $\displaystyle \frac{84}{85}$ C $\displaystyle \frac{36}{85}$ D $\displaystyle \frac{13}{85}$

Answer and solution

Answer: (A) $\displaystyle \frac{77}{85}$

$\displaystyle \cos A = \frac{3}{5}$ , $\displaystyle \sin B = \frac{8}{17}$

$\displaystyle \cos(A - B) = \cos A\cos B + \sin A\sin B = \frac{45}{85} + \frac{32}{85} = \frac{77}{85}$
$\tan 105^\circ =$
A $\sqrt{3} - 2$ B $2 + \sqrt{3}$ C $-(2 + \sqrt{3})$ D $2 - \sqrt{3}$

Answer and solution

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

$\displaystyle \tan(60^\circ + 45^\circ) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = -\frac{(\sqrt{3} + 1)^2}{2} = -(2 + \sqrt{3})$
$\sin^2(45^\circ + \theta) - \sin^2(45^\circ - \theta) =$
A $2\operatorname{sin}\theta$ B $\operatorname{cos} 2\theta$ C $\operatorname{sin} 2\theta$ D $\operatorname{sin}^2\theta$

Answer and solution

Answer: (C) $\operatorname{sin} 2\theta$

$= \sin 90^\circ \sin 2\theta = \sin 2\theta$
$\tan 10^\circ\tan 20^\circ + \tan 20^\circ\tan 60^\circ + \tan 60^\circ\tan 10^\circ =$
A $0$ B $3$ C $\sqrt{3}$ D $1$

Answer and solution

Answer: (D) $1$

$10^\circ + 20^\circ + 60^\circ = 90^\circ$

So $\Sigma\tan A\tan B = 1$
If $A + B = 225^\circ$ , then $(1 + \tan A)(1 + \tan B) =$
A $0$ B $1$ C $2$ D $-2$

Answer and solution

Answer: (C) $2$

$\tan(A + B) = \tan 225^\circ = 1$

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

Answer and solution

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

$\displaystyle \frac{2\tan A}{1 - \tan^2 A} = \tan 2A$

$\displaystyle = \tan 30^\circ = \frac{1}{\sqrt{3}}$
$\displaystyle \frac{3\tan 10^\circ - \tan^3 10^\circ}{1 - 3\tan^2 10^\circ} =$
A $\displaystyle \frac{1}{\sqrt{3}}$ B $\operatorname{tan} 10^\circ$ C $\sqrt{3}$ D $1$

Answer and solution

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

$\displaystyle \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A} = \tan 3A$

$\displaystyle = \tan 30^\circ = \frac{1}{\sqrt{3}}$
$\sin 36^\circ\cos 54^\circ =$
A $\displaystyle \frac{5 + \sqrt{5}}{8}$ B $\displaystyle \frac{1}{4}$ C $\displaystyle \frac{5 - \sqrt{5}}{8}$ D $\displaystyle \frac{3 - \sqrt{5}}{8}$

Answer and solution

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

$\cos 54^\circ = \sin 36^\circ$ , so G.E $= \sin^2 36^\circ$

$\displaystyle = \frac{10 - 2\sqrt{5}}{16} = \frac{5 - \sqrt{5}}{8}$
$\tan 15^\circ(1 + \sec 30^\circ) =$
A $\displaystyle \frac{1}{\sqrt{3}}$ B $\sqrt{3}$ C $2 - \sqrt{3}$ D $1$

Answer and solution

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

$\displaystyle \tan\frac{\theta}{2}(1 + \sec\theta) = \tan\theta$

$\displaystyle = \tan 30^\circ = \frac{1}{\sqrt{3}}$
$\displaystyle \cos\frac{\pi}{11} + \cos\frac{3\pi}{11} + \cos\frac{5\pi}{11} + \cos\frac{7\pi}{11} + \cos\frac{9\pi}{11} =$
A $\displaystyle \frac{5}{2}$ B $\displaystyle \frac{1}{2}$ C $0$ D $\displaystyle -\frac{1}{2}$

Answer and solution

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

$n = 5$ , $\displaystyle d = \frac{2\pi}{11}$ : sum $\displaystyle = \frac{\cos\frac{5\pi}{11}\sin\frac{5\pi}{11}}{\sin\frac{\pi}{11}} = \frac{\sin\frac{10\pi}{11}}{2\sin\frac{\pi}{11}} = \frac{1}{2}$
$\tan 12^\circ + \tan 72^\circ - \tan 48^\circ =$
A $\operatorname{tan} 36^\circ$ B $3\operatorname{cot} 36^\circ$ C $3\operatorname{tan} 12^\circ$ D $3\operatorname{tan} 36^\circ$

Answer and solution

Answer: (D) $3\operatorname{tan} 36^\circ$

$\tan\theta + \tan(60^\circ + \theta) - \tan(60^\circ - \theta) = 3\tan 3\theta$ with $\theta = 12^\circ$
If $0^\circ < A < 90^\circ$ , then $\sqrt{1 + \sin 2A} =$
A $\operatorname{cos} A - \operatorname{sin} A$ B $\operatorname{sin} A - \operatorname{cos} A$ C $\operatorname{cos} A + \operatorname{sin} A$ D $1 + \operatorname{sin} A$

Answer and solution

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

$1 + \sin 2A = (\cos A + \sin A)^2$

$\cos A + \sin A > 0$ for acute $A$
$\sin 105^\circ + \cos 105^\circ =$
A $\displaystyle \frac{\sqrt{3}}{\sqrt{2}}$ B $\displaystyle \frac{1}{\sqrt{2}}$ C $\displaystyle -\frac{1}{\sqrt{2}}$ D $0$

Answer and solution

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

$\cos 105^\circ = \sin(-15^\circ)$

$\displaystyle \sin 105^\circ + \sin(-15^\circ) = 2\sin 45^\circ\cos 60^\circ = \frac{1}{\sqrt{2}}$
The period of $|\sin 2x|$ is
A $\pi$ B $\displaystyle \frac{\pi}{2}$ C $2\pi$ D $\displaystyle \frac{\pi}{4}$

Answer and solution

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

Period of $|\sin x|$ is $\pi$ , so period of $|\sin 2x|$ is $\displaystyle \frac{\pi}{2}$

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MCQ test, Set 7