Question paper, Set 10

$\displaystyle \sec\theta = \frac{\text{hyp}}{\text{adj}} = \frac{65}{56}$: take hyp $= 65$, adj $= 56$

opp $= \sqrt{65^2 - 56^2} = \sqrt{1089} = 33$

$\displaystyle \tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{33}{56}$, $\displaystyle \csc\theta = \frac{\text{hyp}}{\text{opp}} = \frac{65}{33}$

Answer: $\displaystyle \tan\theta = \frac{33}{56}$, $\displaystyle \csc\theta = \frac{65}{33}$

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

$\displaystyle 165^\circ = 165 \times \frac{\pi}{180} = \frac{11\pi}{12}$

Answer: $\displaystyle \frac{11\pi}{12}$

$\displaystyle = (\sqrt{3})^2 - 2\left(\frac{1}{2}\right)^2 - \frac{3}{4}(\sqrt{2})^2$

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

$= 1$

Answer: $1$

Numerator: $\csc(180^\circ + \theta) = -\csc\theta$, $\csc(270^\circ + \theta) = -\sec\theta$, $\tan(90^\circ + \theta) = -\cot\theta$

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

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

Answer: $1$

$\displaystyle \frac{17\pi}{4} = 17 \times 45^\circ = 765^\circ$

$\sec 765^\circ = \sec(720^\circ + 45^\circ) = \sec 45^\circ$

$\sec 45^\circ = \sqrt{2}$

Answer: $\sqrt{2}$

$-1 \le \sin x \le 1$

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

$\Rightarrow -1 \le 1 - 2\sin x \le 3$

Range $= [-1, 3]$

Answer: $[-1, 3]$

$3 = 3^c \approx 171^\circ$

$\sin 171^\circ = \sin(180^\circ - 9^\circ) = \sin 9^\circ$

$3^\circ < 9^\circ \Rightarrow \sin 3^\circ < \sin 3$

Answer: $\sin 3$ is greater

$\cos^2 15^\circ + \cos^2 75^\circ = \cos^2 15^\circ + \sin^2 15^\circ = 1$; similarly for $(25^\circ, 65^\circ)$, $(35^\circ, 55^\circ)$

G.E $\displaystyle = 3(1) + \cos^2 45^\circ = 3 + \frac{1}{2} = \frac{7}{2}$

Answer: $\displaystyle \frac{7}{2}$

Consider $9\cos^2\theta$ and $\sec^2\theta$; both are +ve

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

$\Rightarrow 9\cos^2\theta + \sec^2\theta \ge 6$; equality when $\displaystyle \cos^2\theta = \frac{1}{3}$

Least value $= 6$

Answer: $6$

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

$\displaystyle \cos B = \frac{24}{25} \Rightarrow \sin B = \frac{7}{25}$ (B acute)

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

$\displaystyle = \frac{5}{13} \cdot \frac{24}{25} + \frac{12}{13} \cdot \frac{7}{25} = \frac{24}{65} + \frac{84}{325} = \frac{204}{325}$

Answer: $\displaystyle \frac{204}{325}$

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

$\displaystyle \sin(A + B) = \frac{3}{5} \Rightarrow \cos(A + B) = \frac{4}{5}$, $\displaystyle \tan(A + B) = \frac{3}{4}$

$\displaystyle \sin(A - B) = -\frac{8}{17} \Rightarrow \cos(A - B) = \frac{15}{17}$, $\displaystyle \tan(A - B) = -\frac{8}{15}$

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

$\displaystyle = \frac{\frac{3}{4} - \frac{8}{15}}{1 + \frac{3}{4} \cdot \frac{8}{15}} = \frac{\frac{13}{60}}{\frac{7}{5}}$

$\displaystyle = \frac{13}{84}$

Answer: $\displaystyle \frac{13}{84}$

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

$= \cos 60^\circ \cos 36^\circ$

$\displaystyle = \frac{1}{2} \cdot \frac{\sqrt{5}+1}{4}$

$\displaystyle = \frac{\sqrt{5}+1}{8}$

Answer: $\displaystyle \frac{\sqrt{5}+1}{8}$

In $\triangle ABC$: $\tan A + \tan B + \tan C = \tan A\tan B\tan C$

$1 + 4 + \tan C = 4\tan C$

$\displaystyle (4 - 1)\tan C = 5 \Rightarrow \tan C = \frac{5}{3}$

Answer: $\displaystyle \frac{5}{3}$

$\displaystyle \sin A = \frac{11}{61} \Rightarrow \cos A = \frac{60}{61}$, $\displaystyle \tan A = \frac{11}{60}$ ($A$ acute)

$\displaystyle \sin 2A = 2\sin A\cos A = 2 \cdot \frac{11}{61} \cdot \frac{60}{61} = \frac{1320}{3721}$

$\displaystyle \cos 2A = 1 - 2\sin^2 A = 1 - 2 \cdot \frac{121}{3721} = \frac{3479}{3721}$

Answer: $\displaystyle \sin 2A = \frac{1320}{3721}$, $\displaystyle \cos 2A = \frac{3479}{3721}$

Given $\displaystyle \sin\alpha = \frac{9}{41}$ and $\alpha \in \text{II}_Q$

Right triangle with sides $9$, $40$, $41$; with the signs of $\text{II}_Q$: $\displaystyle \sin\alpha = \frac{9}{41}$, $\displaystyle \cos\alpha = -\frac{40}{41}$, $\displaystyle \tan\alpha = -\frac{9}{40}$

$\displaystyle \cos 3\alpha = 4\cos^3\alpha - 3\cos\alpha = 4\left(-\frac{40}{41}\right)^3 - 3\left(-\frac{40}{41}\right) = -\frac{256000}{68921} + \frac{120}{41} = -\frac{54280}{68921}$

$\displaystyle \tan 2\alpha = \frac{2\tan\alpha}{1 - \tan^2\alpha} = \frac{-\frac{9}{20}}{1 - \frac{81}{1600}} = \frac{-\frac{9}{20}}{\frac{1519}{1600}} = -\frac{720}{1519}$

Answer: $\displaystyle \cos 3\alpha = -\frac{54280}{68921}$, $\displaystyle \tan 2\alpha = -\frac{720}{1519}$

Angles in A.P.: $\displaystyle a_1 = \frac{2\pi}{9}$, $\displaystyle a_n = \frac{8\pi}{9}$, $\displaystyle d = \frac{2\pi}{9}$, $n = 4$

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

$\displaystyle = \frac{\cos\frac{5\pi}{9}\sin\frac{4\pi}{9}}{\sin\frac{\pi}{9}}$, and $\displaystyle \cos\frac{5\pi}{9} = -\cos\frac{4\pi}{9}$

$\displaystyle = -\frac{\sin\frac{8\pi}{9}}{2\sin\frac{\pi}{9}} = -\frac{1}{2}$

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

With $t = \tan\theta$: $\displaystyle \tan(60^\circ + \theta)\tan(60^\circ - \theta) = \frac{3 - t^2}{1 - 3t^2}$

$\displaystyle \tan(60^\circ + \theta) - \tan(60^\circ - \theta) = \frac{8t}{1 - 3t^2}$, so $\displaystyle t[\tan(60^\circ - \theta) - \tan(60^\circ + \theta)] = -\frac{8t^2}{1 - 3t^2}$

LHS $\displaystyle = \frac{3 - t^2 - 8t^2}{1 - 3t^2} = \frac{3(1 - 3t^2)}{1 - 3t^2} = 3$ = RHS

Put $\theta = 5^\circ$: the value is $3$

Answer: $3$

$\displaystyle \frac{x}{y} = \frac{\sin A}{\sin B} \Rightarrow \frac{x + y}{x - y} = \frac{\sin A + \sin B}{\sin A - \sin B} = \frac{\tan\frac{A + B}{2}}{\tan\frac{A - B}{2}}$

$\displaystyle = \frac{\tan 75^\circ}{\tan 30^\circ} = (2 + \sqrt{3})\sqrt{3} = 2\sqrt{3} + 3$

Answer: $3 + 2\sqrt{3}$

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

$\displaystyle \sin(S - C) - \sin S = 2\cos\frac{2S - C}{2}\sin\left(-\frac{C}{2}\right) = -2\cos\frac{A + B}{2}\sin\frac{C}{2}$

LHS $\displaystyle = 2\sin\frac{C}{2}\left[\cos\frac{A - B}{2} - \cos\frac{A + B}{2}\right] = 2\sin\frac{C}{2} \cdot 2\sin\frac{A}{2}\sin\frac{B}{2}$

= RHS

Answer: Proved: LHS $\displaystyle = 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$

Period of $\displaystyle \sin 5x = \frac{2\pi}{5}$, period of $\displaystyle \tan 3x = \frac{\pi}{3}$

Period $=$ L.C.M$\displaystyle \left\{\frac{2\pi}{5}, \frac{\pi}{3}\right\} = \frac{\text{L.C.M}\{2\pi, \pi\}}{\text{G.C.D}\{5, 3\}} = 2\pi$

Answer: $2\pi$