Hypotenuse = 1170. $\sin \theta = 12/13$. $\cos \theta = \sqrt{1-(12/13)^2} = \sqrt{25/169} = 5/13$. Horizontal distance = Hypotenuse $\cdot \cos \theta = 1170 \cdot (5/13) = 90 \cdot 5 = 450$ meters.

Number of terms: $1, 3, 5, \dots, 89$ is 45 terms. $\sin^2 1^\circ + \sin^2 89^\circ = \sin^2 1^\circ + \cos^2 1^\circ = 1$. Thus, there will be 22 pairs and $\sin^2 45^\circ$ will remain. $22 + (1/\sqrt{2})^2 = 22 + 1/2 = 22.5$

$\angle A=90^\circ$, $a=BC=25, b=AC=7$. $AB = \sqrt{BC^2 - AC^2} = \sqrt{25^2-7^2} = \sqrt{625-49} = \sqrt{576} = 24$. $\tan B = AC/AB = 7/24$.

By Componendo and Dividendo: $\frac{(\sec\theta + \tan\theta) + (\sec\theta - \tan\theta)}{(\sec\theta + \tan\theta) - (\sec\theta - \tan\theta)} = \frac{(2+\sqrt{3}) + (2-\sqrt{3})}{(2+\sqrt{3}) - (2-\sqrt{3})} \implies \frac{2\sec\theta}{2\tan\theta} = \frac{4}{2\sqrt{3}} \implies \frac{1}{\sin\theta} = \frac{2}{\sqrt{3}} \implies \sin\theta = \frac{\sqrt{3}}{2} = \sin 60^\circ \implies \theta = 60^\circ = \frac{\pi}{3}$.

$(\sin A + \cos A) \sec A = \sin A \cdot \sec A + \cos A \cdot \sec A = \tan A + 1 = \frac{5}{12} + 1 = \frac{17}{12}$

Expression $= \sqrt{(\sin^2 \theta + \cos^2 \theta) + 1} = \sqrt{1+1} = \sqrt{2}$.

$\tan 30^\circ = \frac{h}{x} \Rightarrow x = h\sqrt{3}$. $\tan 15^\circ = \frac{h}{x+100} = \frac{h}{h\sqrt{3}+100}$. $2-\sqrt{3} = \frac{h}{h\sqrt{3}+100} \Rightarrow (2-\sqrt{3})(h\sqrt{3}+100) = h \Rightarrow 2h\sqrt{3} + 200 - 3h - 100\sqrt{3} = h \Rightarrow h(2\sqrt{3}-4) = 100(\sqrt{3}-2) \Rightarrow h = \frac{100(\sqrt{3}-2)}{2(\sqrt{3}-2)} = 50$ meters.

Square both sides: $(\tan \theta + \cot \theta)^2 = 6^2 \Rightarrow \tan^2 \theta + \cot^2 \theta + 2 = 36 \Rightarrow \tan^2 \theta + \cot^2 \theta = 34$. Answer is 34.

$\cos \theta = \frac{\text{base}}{\text{hyp}} = \frac{a}{b}$. Perpendicular $= \sqrt{b^2 - a^2}$. $\tan \theta = \frac{\text{perp}}{\text{base}} = \frac{\sqrt{b^2 - a^2}}{a}$.

Rationalizing: $\sqrt{\frac{(\sec \theta - 1)(\sec \theta - 1)}{(\sec \theta + 1)(\sec \theta - 1)}} = \sqrt{\frac{(\sec \theta - 1)^2}{\sec^2 \theta - 1}} = \sqrt{\frac{(\sec \theta - 1)^2}{\tan^2 \theta}} = \frac{\sec \theta - 1}{\tan \theta} = \frac{1/\cos \theta - 1}{\sin \theta / \cos \theta} = \frac{1-\cos \theta}{\sin \theta} = \csc \theta - \cot \theta$.

$0 < \theta \le 90^\circ$ ($\theta$ is acute angle) and $\sin\theta = \cos\theta$. $\Rightarrow \theta = 45^\circ$. Now, $2\cot^2 45^\circ + \cos^2 45^\circ + 1 = 2(1)^2 + (\frac{1}{\sqrt{2}})^2 + 1 = 2 + 0.5 + 1 = 3.5 = \frac{7}{2}$.

$\tan X = (1/\sqrt{2}) \cdot (1/\sqrt{2}) + 1/2 = 1/2 + 1/2 = 1$. $\tan X = 1 \implies X = 45^{\circ}$.

Let the height of the pole be h meters. From the figure: $\tan 30^{\circ} = \frac{90-h}{x} \implies x = \sqrt{3}(90-h)$ ... (i). And $\tan 60^{\circ} = \frac{90}{x} \implies x = \frac{90}{\sqrt{3}}$. From (i) $\sqrt{3}(90-h) = \frac{90}{\sqrt{3}} \implies 90-h = 30 \implies h = 60$ meters.

$\sec\theta + \tan\theta = x$. We know $\sec^2\theta - \tan^2\theta = 1 \Rightarrow (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1 \Rightarrow \sec\theta - \tan\theta = \frac{1}{x}$. Now, $(\sec\theta + \tan\theta) - (\sec\theta - \tan\theta) = x - \frac{1}{x} \Rightarrow 2\tan\theta = \frac{x^2-1}{x} \Rightarrow \tan\theta = \frac{x^2-1}{2x}$.

Let height of the tree be $H$ meters. In $\triangle ABC$ (where wall height is 24m), $\tan 60^\circ = \frac{H-24}{x} \Rightarrow H-24 = x\sqrt{3}$. In $\triangle CED$, $\tan 30^\circ = \frac{24}{x} \Rightarrow x = 24\sqrt{3}$. Thus $H-24 = (24\sqrt{3})\sqrt{3} = 24(3) = 72 \Rightarrow H = 96$ meters. Ratio $\frac{H}{x} = \frac{96}{24\sqrt{3}} = \frac{4}{\sqrt{3}}$. (This simplifies to $4\sqrt{3}:3$, but book option is $2\sqrt{3}:1$).

Let the height of the pole be h, shadow = $\sqrt{3}h$. $\tan \theta = \frac{h}{\sqrt{3}h} = \frac{1}{\sqrt{3}}$. $\theta = 30^{\circ}$.

Total terms = $(89-1)/2 + 1 = 45$. Term $\sin^2 45^{\circ}$ will be in the middle. The remaining 44 terms ($\sin^2 \theta + \sin^2 (90-\theta) = 1$) will make 22 pairs. Total = $22 + \sin^2 45^{\circ} = 22 + 1/2 = 22.5$.

$\cot A = 3/4$. Dividing numerator and denominator by $\sin A$: $\frac{1 - 4\cot A}{1 + 4\cot A} = \frac{1 - 4(3/4)}{1 + 4(3/4)} = \frac{1 - 3}{1 + 3} = -2/4 = -1/2$.

$\sin \theta + \cos \theta = \sqrt{2} \cos \theta \implies \sin \theta = (\sqrt{2} - 1) \cos \theta \implies \tan \theta = \sqrt{2} - 1 \implies \cot \theta = \frac{1}{\sqrt{2}-1} = \sqrt{2} + 1$.

$\tan 60^\circ = \text{wall height} / 1.5 \Rightarrow \sqrt{3} = h / 1.5 \Rightarrow h = 1.5\sqrt{3}$ meters.