Question paper, Set 2

$\displaystyle \tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{5}{12}$: take opp $= 5$, adj $= 12$

hyp $= \sqrt{5^2 + 12^2} = \sqrt{169} = 13$

$\displaystyle \sin\theta = \frac{\text{opp}}{\text{hyp}} = \frac{5}{13}$, $\displaystyle \sec\theta = \frac{\text{hyp}}{\text{adj}} = \frac{13}{12}$

Answer: $\displaystyle \sin\theta = \frac{5}{13}$, $\displaystyle \sec\theta = \frac{13}{12}$

$\pi = 180^\circ$

$\displaystyle \frac{7\pi}{9} = \frac{7}{9} \times 180^\circ = 140^\circ$

Answer: $140^\circ$

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

$\displaystyle = \frac{3}{4} + \frac{1}{2} + \frac{1}{3} = \frac{9 + 6 + 4}{12}$

$\displaystyle = \frac{19}{12}$

Answer: $\displaystyle \frac{19}{12}$

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

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

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

Answer: $1$

$\cos 1500^\circ = \cos(1440^\circ + 60^\circ) = \cos 60^\circ$

$\displaystyle \cos 60^\circ = \frac{1}{2}$

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

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

$\Rightarrow -5 \le 5\cos x \le 5$

Range $= [-5, 5]$

Answer: $[-5, 5]$

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

From the graph, $\cos x$ decreases from $1$ to $0$ as $x$ goes from $0^\circ$ to $90^\circ$

$1^\circ < 57^\circ \Rightarrow \cos 1^\circ > \cos 1$

Answer: $\cos 1^\circ$ is greater

In a cyclic quadrilateral $A + C = 180^\circ$ and $B + D = 180^\circ$

$\tan C = \tan(180^\circ - A) = -\tan A$, $\tan D = \tan(180^\circ - B) = -\tan B$

G.E $= \tan A + \tan B - \tan A - \tan B = 0$

Answer: $0$

Consider $9\tan^2\theta$ and $4\cot^2\theta$; both are +ve

A.M $\ge$ G.M: $\displaystyle \frac{9\tan^2\theta + 4\cot^2\theta}{2} \ge \sqrt{9\tan^2\theta \cdot 4\cot^2\theta} = \sqrt{36} = 6$

$\Rightarrow 9\tan^2\theta + 4\cot^2\theta \ge 12$; equality when $\displaystyle \tan^2\theta = \frac{2}{3}$

Least value $= 12$

Answer: $12$

$\displaystyle \cos A = \frac{15}{17} \Rightarrow \sin A = \frac{8}{17}$ (A acute)

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

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

$\displaystyle = \frac{15}{17} \cdot \frac{21}{29} - \frac{8}{17} \cdot \frac{20}{29} = \frac{315}{493} - \frac{160}{493} = \frac{155}{493}$

Answer: $\displaystyle \frac{155}{493}$

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

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

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

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

$\displaystyle = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \cdot \frac{5}{12}} = \frac{\frac{7}{6}}{\frac{11}{16}}$

$\displaystyle = \frac{56}{33}$

Answer: $\displaystyle \frac{56}{33}$

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

$= \sin 90^\circ \sin 60^\circ$

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

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

Answer: $\displaystyle \frac{\sqrt{3}}{2}$

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

$\displaystyle \frac{1}{6} + \left(\frac{1}{2} + \frac{1}{3}\right)\tan C = 1$

$\displaystyle \frac{5}{6}\tan C = \frac{5}{6} \Rightarrow \tan C = 1$

Answer: $1$

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

$\displaystyle \sin 2A = 2\sin A\cos A = 2 \cdot \frac{5}{13} \cdot \frac{12}{13} = \frac{120}{169}$

$\displaystyle \tan 2A = \frac{2\tan A}{1 - \tan^2 A} = \frac{\frac{5}{6}}{\frac{119}{144}} = \frac{120}{119}$

Answer: $\displaystyle \sin 2A = \frac{120}{169}$, $\displaystyle \tan 2A = \frac{120}{119}$

Given $\displaystyle \cos\alpha = -\frac{3}{5}$ and $\alpha \in \text{III}_Q$

Right triangle with sides $4$, $3$, $5$; with the signs of $\text{III}_Q$: $\displaystyle \sin\alpha = -\frac{4}{5}$, $\displaystyle \cos\alpha = -\frac{3}{5}$, $\displaystyle \tan\alpha = \frac{4}{3}$

$\displaystyle \cos 3\alpha = 4\cos^3\alpha - 3\cos\alpha = 4\left(-\frac{3}{5}\right)^3 - 3\left(-\frac{3}{5}\right) = -\frac{108}{125} + \frac{9}{5} = \frac{117}{125}$

$\displaystyle \sin 2\alpha = 2\sin\alpha\cos\alpha = 2\left(-\frac{4}{5}\right)\left(-\frac{3}{5}\right) = \frac{24}{25}$

Answer: $\displaystyle \cos 3\alpha = \frac{117}{125}$, $\displaystyle \sin 2\alpha = \frac{24}{25}$

Angles in A.P.: $\displaystyle a_1 = \frac{\pi}{9}$, $\displaystyle a_n = \frac{7\pi}{9}$, $\displaystyle d = \frac{2\pi}{9}$, $n = 4$

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

$\displaystyle = \frac{\cos\frac{4\pi}{9}\sin\frac{4\pi}{9}}{\sin\frac{\pi}{9}} = \frac{\sin\frac{8\pi}{9}}{2\sin\frac{\pi}{9}}$

$\displaystyle \sin\frac{8\pi}{9} = \sin\left(\pi - \frac{\pi}{9}\right) = \sin\frac{\pi}{9}$, so the sum $\displaystyle = \frac{1}{2}$

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

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

LHS $\displaystyle = t\cdot\frac{3 - t^2}{1 - 3t^2} = \frac{3t - t^3}{1 - 3t^2} = \tan 3\theta =$ RHS

Put $\theta = 20^\circ$: $\tan 20^\circ\tan 40^\circ\tan 80^\circ = \tan 60^\circ = \sqrt{3}$

Answer: $\sqrt{3}$

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

$\displaystyle = \frac{\tan 60^\circ}{\tan 45^\circ} = \sqrt{3}$

Answer: $\sqrt{3}$

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

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

$= -1 - 2\cos C[\cos(A - B) - \cos C] = -1 - 2\cos C[\cos(A - B) + \cos(A + B)]$

$= -1 - 2\cos C \cdot 2\cos A\cos B = -1 - 4\cos A\cos B\cos C$

= RHS

Answer: Proved: LHS $= -1 - 4\cos A\cos B\cos C$

Period of $\tan(ax + b) + k$ is $\displaystyle \frac{\pi}{|a|}$

$\displaystyle = \frac{\pi}{4}$

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