Advanced Trigonometry: MCQ test, Set 9

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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 $\displaystyle \csc\theta = \frac{41}{40}$ and $\theta$ is acute, then $\sec\theta =$
A $\displaystyle \frac{9}{41}$ B $\displaystyle \frac{40}{41}$ C $\displaystyle \frac{41}{9}$ D $\displaystyle \frac{40}{9}$

Answer and solution

Answer: (C) $\displaystyle \frac{41}{9}$

$\displaystyle \csc\theta = \frac{\text{hyp}}{\text{opp}} = \frac{41}{40}$ : take hyp $= 41$ , opp $= 40$

adj $= \sqrt{41^2 - 40^2} = \sqrt{81} = 9$

$\displaystyle \sec\theta = \frac{\text{hyp}}{\text{adj}} = \frac{41}{9}$
$1440^\circ$ in radians is
A $16\pi$ B $8\pi$ C $6\pi$ D $4\pi$

Answer and solution

Answer: (B) $8\pi$

$1440^\circ = 4 \times 360^\circ = 4 \times 2\pi = 8\pi$
The sign of $\csc 200^\circ\tan 120^\circ\cos 80^\circ$ is
A positive B not defined C zero D negative

Answer and solution

Answer: (A) positive

$200^\circ \in \text{III}_Q \Rightarrow \csc 200^\circ$ is -ve

$120^\circ \in \text{II}_Q \Rightarrow \tan 120^\circ$ is -ve

$80^\circ \in \text{I}_Q \Rightarrow \cos 80^\circ$ is +ve

So the product is positive
$\cos 0^\circ + \sin 90^\circ + \tan 45^\circ =$
A $0$ B $3$ C $1$ D $2$

Answer and solution

Answer: (B) $3$

$= 1 + 1 + 1$

$= 3$
$\operatorname{cos}(180^\circ + \theta) =$
A $-\operatorname{sin}\theta$ B $\operatorname{cos}\theta$ C $\operatorname{sin}\theta$ D $-\operatorname{cos}\theta$

Answer and solution

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

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

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

$\therefore \cos(180^\circ + \theta) = -\cos\theta$
$\displaystyle \tan \frac{35\pi}{4} =$
A $1$ B $-1$ C $\displaystyle \frac{1}{2}$ D $\displaystyle \frac{\sqrt{3}}{2}$

Answer and solution

Answer: (B) $-1$

$\displaystyle \frac{35\pi}{4} = 35 \times 45^\circ = 1575^\circ$

$\tan 1575^\circ = \tan(1440^\circ + 135^\circ) = \tan 135^\circ$

$135^\circ \in \text{II}_Q$ , where $\tan$ is -ve

$\tan 135^\circ = \tan(180^\circ - 45^\circ) = -\tan 45^\circ = -1$
The greatest value of $3 - 4\cos x$ is
A $-1$ B $4$ C $7$ D $3$

Answer and solution

Answer: (C) $7$

$-1 \le \cos x \le 1 \Rightarrow -4 \le -4\cos x \le 4$

$\Rightarrow -1 \le 3 - 4\cos x \le 7$
For the radian angles $4$ , $5$ , $6$ , the cosines satisfy
A $\operatorname{cos} 4 > \operatorname{cos} 5 > \operatorname{cos} 6$ B $\operatorname{cos} 6 > \operatorname{cos} 4 > \operatorname{cos} 5$ C $\operatorname{cos} 6 > \operatorname{cos} 5 > \operatorname{cos} 4$ D $\operatorname{cos} 5 > \operatorname{cos} 6 > \operatorname{cos} 4$

Answer and solution

Answer: (C) $\operatorname{cos} 6 > \operatorname{cos} 5 > \operatorname{cos} 4$

$\cos 229^\circ = -\cos 49^\circ < 0$ , $\cos 286^\circ = \cos 74^\circ$ , $\cos 343^\circ = \cos 17^\circ$

$\cos 17^\circ > \cos 74^\circ > -\cos 49^\circ$
If $A + B = 90^\circ$ , then $\sin^2 A + \sin^2 B =$
A $2\operatorname{sin}^2 A$ B $0$ C $\displaystyle \frac{1}{2}$ D $1$

Answer and solution

Answer: (D) $1$

$\sin B = \sin(90^\circ - A) = \cos A$

$\sin^2 A + \cos^2 A = 1$
The value of $\sin^6\theta + \cos^6\theta$ at $\theta = 30^\circ$ is
A $\displaystyle \frac{5}{8}$ B $\displaystyle \frac{7}{16}$ C $\displaystyle \frac{1}{2}$ D $1$

Answer and solution

Answer: (B) $\displaystyle \frac{7}{16}$

$\displaystyle = 1 - 3\sin^2 30^\circ\cos^2 30^\circ = 1 - 3\cdot\frac{1}{4}\cdot\frac{3}{4} = 1 - \frac{9}{16} = \frac{7}{16}$
The least value of $9\sin^2\theta + \csc^2\theta$ is
A $6$ B $2$ C $3$ D $10$

Answer and solution

Answer: (A) $6$

Consider $9\sin^2\theta$ and $\csc^2\theta$ ; both are +ve

A.M $\ge$ G.M: $\displaystyle \frac{9\sin^2\theta + \csc^2\theta}{2} \ge \sqrt{9\sin^2\theta \cdot \csc^2\theta} = \sqrt{9} = 3$

$\Rightarrow 9\sin^2\theta + \csc^2\theta \ge 6$ ; equality when $\displaystyle \sin^2\theta = \frac{1}{3}$
$\sin 15^\circ + \cos 15^\circ =$
A $1$ B $\sqrt{2}$ C $\displaystyle \frac{1}{\sqrt{2}}$ D $\displaystyle \frac{\sqrt{3}}{\sqrt{2}}$

Answer and solution

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

$\displaystyle \sin 15^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}}$ , $\displaystyle \cos 15^\circ = \frac{\sqrt{3}+1}{2\sqrt{2}}$

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

Answer and solution

Answer: (B) $1$

$\sqrt{3} = \tan 60^\circ$ : G.E $= \tan(60^\circ - 15^\circ) = \tan 45^\circ = 1$
If $\displaystyle \sin A = \frac{1}{2}$ and $\displaystyle \sin B = \frac{1}{3}$ , then $\sin(A + B)\sin(A - B) =$
A $\displaystyle -\frac{5}{36}$ B $\displaystyle \frac{5}{36}$ C $\displaystyle \frac{13}{36}$ D $\displaystyle \frac{1}{6}$

Answer and solution

Answer: (B) $\displaystyle \frac{5}{36}$

$\sin(A + B)\sin(A - B) = \sin^2 A - \sin^2 B$

$\displaystyle = \frac{1}{4} - \frac{1}{9} = \frac{5}{36}$
In $\triangle ABC$ , if $\cot A = 1$ and $\cot B = 2$ , then $\cot C =$
A $\displaystyle -\frac{1}{3}$ B $3$ C $\displaystyle \frac{1}{3}$ D $-3$

Answer and solution

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

$\Sigma\cot A\cot B = 1$ : $2 + 3\cot C = 1$

$\displaystyle \cot C = -\frac{1}{3}$
$\tan 110^\circ + \tan 115^\circ + \tan 110^\circ\tan 115^\circ =$
A $-1$ B $2$ C $1$ D $0$

Answer and solution

Answer: (C) $1$

$110^\circ + 115^\circ = 225^\circ$ , $\tan 225^\circ = 1$

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

Answer and solution

Answer: (D) $\displaystyle \frac{3}{5}$

$\displaystyle \sin^2 A = \frac{1 - \cos 2A}{2} = \frac{1 - \frac{7}{25}}{2} = \frac{9}{25}$

$\displaystyle \sin A = \frac{3}{5}$ ( $A$ acute)
If $\displaystyle \tan\frac{A}{2} = \frac{1}{3}$ , then $\sin A =$
A $\displaystyle \frac{3}{5}$ B $\displaystyle \frac{4}{5}$ C $\displaystyle \frac{3}{4}$ D $\displaystyle \frac{2}{3}$

Answer and solution

Answer: (A) $\displaystyle \frac{3}{5}$

$\displaystyle \sin A = \frac{2\tan\frac{A}{2}}{1 + \tan^2\frac{A}{2}} = \frac{2/3}{10/9} = \frac{3}{5}$
$\cos 36^\circ\cos 72^\circ =$
A $\displaystyle \frac{1}{2}$ B $\displaystyle \frac{1}{8}$ C $\displaystyle \frac{1}{4}$ D $\displaystyle \frac{\sqrt{5}}{4}$

Answer and solution

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

$\cos 72^\circ = \sin 18^\circ$

$\displaystyle \frac{\sqrt{5}+1}{4} \cdot \frac{\sqrt{5}-1}{4} = \frac{5 - 1}{16} = \frac{1}{4}$
$\cot x - \tan x - 2\tan 2x =$
A $4\operatorname{tan} 4x$ B $8\operatorname{cot} 8x$ C $4\operatorname{cot} 4x$ D $2\operatorname{cot} 2x$

Answer and solution

Answer: (C) $4\operatorname{cot} 4x$

$\cot x - \tan x = 2\cot 2x$

$2(\cot 2x - \tan 2x) = 2 \cdot 2\cot 4x = 4\cot 4x$
$\displaystyle \cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} =$
A $\displaystyle -\frac{1}{2}$ B $0$ C $-1$ D $\displaystyle \frac{1}{2}$

Answer and solution

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

$n = 2$ , $\displaystyle d = \frac{2\pi}{5}$ : sum $\displaystyle = \frac{\cos\frac{3\pi}{5}\sin\frac{2\pi}{5}}{\sin\frac{\pi}{5}} = -\frac{\sin\frac{4\pi}{5}}{2\sin\frac{\pi}{5}} = -\frac{1}{2}$
$\sin 10^\circ\sin 50^\circ\sin 70^\circ =$
A $\displaystyle \frac{1}{4}$ B $\displaystyle \frac{1}{16}$ C $\displaystyle \frac{\sqrt{3}}{8}$ D $\displaystyle \frac{1}{8}$

Answer and solution

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

$\theta = 10^\circ$ : $\displaystyle = \frac{1}{4}\sin 30^\circ = \frac{1}{8}$
If $\displaystyle \frac{x}{y} = \frac{\tan A}{\tan B}$ , then $\displaystyle \frac{x - y}{x + y} =$
A $\displaystyle \frac{\operatorname{sin}(A - B)}{\operatorname{sin}(A + B)}$ B $\displaystyle \frac{\operatorname{cos}(A - B)}{\operatorname{cos}(A + B)}$ C $\displaystyle \frac{\operatorname{tan}(A - B)}{\operatorname{tan}(A + B)}$ D $\displaystyle \frac{\operatorname{sin}(A + B)}{\operatorname{sin}(A - B)}$

Answer and solution

Answer: (A) $\displaystyle \frac{\operatorname{sin}(A - B)}{\operatorname{sin}(A + B)}$

$\displaystyle \frac{x - y}{x + y} = \frac{\tan A - \tan B}{\tan A + \tan B} = \frac{\sin A\cos B - \cos A\sin B}{\sin A\cos B + \cos A\sin B} = \frac{\sin(A - B)}{\sin(A + B)}$
$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ =$
A $\displaystyle \frac{1}{2}$ B $1$ C $2\operatorname{sin} 10^\circ$ D $0$

Answer and solution

Answer: (D) $0$

$\sin 50^\circ + \sin 10^\circ = 2\sin 30^\circ\cos 20^\circ = \cos 20^\circ = \sin 70^\circ$

So G.E $= 0$
The period of $\displaystyle -3\sin\frac{x}{2}$ is
A $2\pi$ B $6\pi$ C $4\pi$ D $\pi$

Answer and solution

Answer: (C) $4\pi$

$-f$ has the same period as $f$ : $\displaystyle \frac{2\pi}{1/2} = 4\pi$

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MCQ test, Set 9