Question paper, Set 6

$\displaystyle \csc\theta = \frac{\text{hyp}}{\text{opp}} = \frac{41}{9}$: take hyp $= 41$, opp $= 9$

adj $= \sqrt{41^2 - 9^2} = \sqrt{1600} = 40$

$\displaystyle \cos\theta = \frac{\text{adj}}{\text{hyp}} = \frac{40}{41}$, $\displaystyle \tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{9}{40}$

Answer: $\displaystyle \cos\theta = \frac{40}{41}$, $\displaystyle \tan\theta = \frac{9}{40}$

$180^\circ = \pi$, so $\displaystyle 1^\circ = \frac{\pi}{180}$

$\displaystyle 252^\circ = 252 \times \frac{\pi}{180} = \frac{7\pi}{5}$

Answer: $\displaystyle \frac{7\pi}{5}$

$\displaystyle \tan^2 30^\circ = \frac{1}{3}$

$\displaystyle = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{2/3}{4/3}$

$\displaystyle = \frac{1}{2}$

Answer: $\displaystyle \frac{1}{2}$

Numerator: $\sec(270^\circ + \theta) = \csc\theta$, $\cos(-\theta) = \cos\theta$, $\csc(90^\circ + \theta) = \sec\theta$

Denominator: $\csc(360^\circ - \theta) = -\csc\theta$, $\sin(90^\circ + \theta) = \cos\theta$, $\sec(180^\circ + \theta) = -\sec\theta$

G.E $\displaystyle = \frac{(\csc\theta)(\cos\theta)(\sec\theta)}{(-\csc\theta)(\cos\theta)(-\sec\theta)} = 1$

Answer: $1$

$\cot 690^\circ = \cot(360^\circ + 330^\circ) = \cot 330^\circ$

$330^\circ \in \text{IV}_Q$, where $\cot$ is -ve

$\cot 330^\circ = \cot(360^\circ - 30^\circ) = -\cot 30^\circ = -\sqrt{3}$

Answer: $-\sqrt{3}$

$-1 \le \cos x \le 1$

$\Rightarrow -3 \le -3\cos x \le 3$

$\Rightarrow 1 \le 4 - 3\cos x \le 7$

Range $= [1, 7]$

Answer: $[1, 7]$

$1 = 1^c \approx 57^\circ$

From the graph, $\tan x$ increases from $0^\circ$ to $90^\circ$

$57^\circ > 1^\circ \Rightarrow \tan 1 > \tan 1^\circ$

Answer: $\tan 1$ is greater

$1^\circ + 179^\circ = 180^\circ$: $\cos 1^\circ + \cos 179^\circ = \cos 1^\circ - \cos 1^\circ = 0$

The terms make 89 such pairs, and $\cos 90^\circ = 0$ is left

G.E $= 0$

Answer: $0$

$\sqrt{1 - \cos^2 200^\circ} = \sqrt{\sin^2 200^\circ} = |\sin 200^\circ|$

$200^\circ \in \text{III}_Q \Rightarrow \sin 200^\circ < 0$

$\therefore |\sin 200^\circ| = -\sin 200^\circ$

Answer: $-\sin 200^\circ$

$\displaystyle \cos A = \frac{3}{5} \Rightarrow \sin A = \frac{4}{5}$ (A acute)

$\displaystyle \sin B = \frac{21}{29} \Rightarrow \cos B = \frac{20}{29}$ (B acute)

$\cos(A + B) = \cos A\cos B - \sin A\sin B$

$\displaystyle = \frac{3}{5} \cdot \frac{20}{29} - \frac{4}{5} \cdot \frac{21}{29} = \frac{12}{29} - \frac{84}{145} = -\frac{24}{145}$

Answer: $\displaystyle -\frac{24}{145}$

$\displaystyle 0 < A < B < \frac{\pi}{4} \Rightarrow 0 < A + B < \frac{\pi}{2}$ and $A - B < 0$

$\displaystyle \tan(A + B) = \frac{21}{20}$

$\displaystyle \sin(A - B) = -\frac{5}{13} \Rightarrow \cos(A - B) = \frac{12}{13}$, $\displaystyle \tan(A - B) = -\frac{5}{12}$

$\displaystyle \tan 2B = \tan[(A + B) - (A - B)] = \frac{\tan(A + B) - \tan(A - B)}{1 + \tan(A + B)\tan(A - B)}$

$\displaystyle = \frac{\frac{21}{20} + \frac{5}{12}}{1 - \frac{21}{20} \cdot \frac{5}{12}} = \frac{\frac{22}{15}}{\frac{9}{16}}$

$\displaystyle = \frac{352}{135}$

Answer: $\displaystyle \frac{352}{135}$

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

$= \cos 60^\circ \cos 30^\circ$

$\displaystyle = \frac{1}{2} \cdot \frac{\sqrt{3}}{2}$

$\displaystyle = \frac{\sqrt{3}}{4}$

Answer: $\displaystyle \frac{\sqrt{3}}{4}$

If $A + B + C = 90^\circ$: $\tan A\tan B + \tan B\tan C + \tan C\tan A = 1$

$\displaystyle \frac{1}{20} + \left(\frac{1}{4} + \frac{1}{5}\right)\tan C = 1$

$\displaystyle \frac{9}{20}\tan C = \frac{19}{20} \Rightarrow \tan C = \frac{19}{9}$

Answer: $\displaystyle \frac{19}{9}$

$\displaystyle \cos A = \frac{40}{41} \Rightarrow \sin A = \frac{9}{41}$, $\displaystyle \tan A = \frac{9}{40}$ ($A$ acute)

$\displaystyle \sin 2A = 2\sin A\cos A = 2 \cdot \frac{9}{41} \cdot \frac{40}{41} = \frac{720}{1681}$

$\displaystyle \tan 2A = \frac{2\tan A}{1 - \tan^2 A} = \frac{\frac{9}{20}}{\frac{1519}{1600}} = \frac{720}{1519}$

Answer: $\displaystyle \sin 2A = \frac{720}{1681}$, $\displaystyle \tan 2A = \frac{720}{1519}$

Given $\displaystyle \tan\alpha = \frac{20}{21}$ and $\alpha \in \text{III}_Q$

Right triangle with sides $20$, $21$, $29$; with the signs of $\text{III}_Q$: $\displaystyle \sin\alpha = -\frac{20}{29}$, $\displaystyle \cos\alpha = -\frac{21}{29}$, $\displaystyle \tan\alpha = \frac{20}{21}$

$\displaystyle \sin 2\alpha = 2\sin\alpha\cos\alpha = 2\left(-\frac{20}{29}\right)\left(-\frac{21}{29}\right) = \frac{840}{841}$

$\displaystyle \tan 3\alpha = \frac{3\tan\alpha - \tan^3\alpha}{1 - 3\tan^2\alpha} = \frac{\frac{20}{7} - \frac{8000}{9261}}{1 - \frac{400}{147}} = -\frac{18460}{15939}$

Answer: $\displaystyle \sin 2\alpha = \frac{840}{841}$, $\displaystyle \tan 3\alpha = -\frac{18460}{15939}$

Angles in A.P.: $\displaystyle a_1 = \frac{\pi}{13}$, $\displaystyle a_n = \frac{11\pi}{13}$, $\displaystyle d = \frac{2\pi}{13}$, $n = 6$

Sum $\displaystyle = \frac{\cos\frac{a_1 + a_n}{2}\,\sin\frac{nd}{2}}{\sin\frac{d}{2}}$

$\displaystyle = \frac{\cos\frac{6\pi}{13}\sin\frac{6\pi}{13}}{\sin\frac{\pi}{13}} = \frac{\sin\frac{12\pi}{13}}{2\sin\frac{\pi}{13}}$

$\displaystyle \sin\frac{12\pi}{13} = \sin\frac{\pi}{13}$, so the sum $\displaystyle = \frac{1}{2}$

Answer: $\displaystyle \frac{1}{2}$

$\displaystyle \cos^2 A + \cos^2 B = 1 + (\cos^2 A - \sin^2 B) = 1 + \cos(A + B)\cos(A - B) = 1 + \frac{1}{2}\cos(A + B)$

$\displaystyle \cos A\cos B = \frac{1}{2}[\cos(A + B) + \cos(A - B)] = \frac{1}{2}\cos(A + B) + \frac{1}{4}$

LHS $\displaystyle = 1 + \frac{1}{2}\cos(A + B) - \frac{1}{2}\cos(A + B) - \frac{1}{4} = \frac{3}{4}$ = RHS

$80^\circ - 20^\circ = 60^\circ$, so the value is $\displaystyle \frac{3}{4}$

Answer: $\displaystyle \frac{3}{4}$

$1 + \sin 100^\circ = \cos^2 50^\circ + \sin^2 50^\circ + 2\sin 50^\circ\cos 50^\circ = (\cos 50^\circ + \sin 50^\circ)^2$

$\sqrt{1 + \sin 100^\circ} = |\cos 50^\circ + \sin 50^\circ| = \cos 50^\circ + \sin 50^\circ$ (both +ve)

Answer: $\cos 50^\circ + \sin 50^\circ$

Given $A + B + C = 180^\circ$, so $\sin(A + B) = \sin C$ and $\cos(A + B) = -\cos C$

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

LHS $= 2 + \cos C\cos(A - B) - \cos^2 C = 2 + \cos C[\cos(A - B) + \cos(A + B)]$

$= 2 + \cos C \cdot 2\cos A\cos B = 2 + 2\cos A\cos B\cos C$

= RHS

Answer: Proved: LHS $= 2 + 2\cos A\cos B\cos C$

$\displaystyle \cos 3x = 4\cos^3 x - 3\cos x \Rightarrow \cos^3 x = \frac{3}{4}\cos x + \frac{1}{4}\cos 3x$

Period $=$ L.C.M$\displaystyle \left\{2\pi, \frac{2\pi}{3}\right\} = \frac{\text{L.C.M}\{2\pi, 2\pi\}}{\text{G.C.D}\{1, 3\}} = 2\pi$

Answer: $2\pi$