1. **Find the value of Sin 105°** We use the angle sum formula: $$\sin(105^\circ) = \sin(60^\circ + 45^\circ) = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ$$ Using known values, $$\sin 60^\circ = \frac{\sqrt{3}}{2}, \cos 45^\circ = \frac{\sqrt{2}}{2}, \cos 60^\circ = \frac{1}{2}, \sin 45^\circ = \frac{\sqrt{2}}{2}$$ Substitute: $$\sin 105^\circ = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$$ 2. **Find equation of straight line with slope -2 and y-intercept -5** Slope-intercept form: $$y = mx + c$$ Given, $$m = -2, c = -5$$ So, $$y = -2x - 5$$ 3. **Find constant term in $\left(2x^2 + \frac{3}{x}\right)^9$** General term of expansion: $$T_{r+1} = \binom{9}{r} (2x^2)^{9-r} \left(\frac{3}{x}\right)^r = \binom{9}{r} 2^{9-r} 3^r x^{2(9-r)-r}$$ Simplify exponent of $x$: $$2(9-r) - r = 18 - 2r - r = 18 - 3r$$ Constant term occurs when exponent of $x$ is zero: $$18 - 3r = 0 \Rightarrow r = 6$$ Substitute $r=6$: $$T_7 = \binom{9}{6} 2^{3} 3^{6}$$ Calculate: $$\binom{9}{6} = 84, \quad 2^3 = 8, \quad 3^6 = 729$$ So, constant term = $$84 \times 8 \times 729 = 490, 992$$ 4. **Find values of $p$ and $q$ so that both roots of $$x^2 + (2p - 4)x - 3 = 0$$ are real and equal** Discriminant for equal roots: $$D = b^2 - 4ac = 0$$ Here, $$a=1, b=2p - 4, c=-3$$ Calculate: $$(2p - 4)^2 - 4(1)(-3) = 0 \Rightarrow (2p - 4)^2 + 12 = 0$$ But $(2p - 4)^2 \geq 0$ and $+12$ makes it positive, so discriminant never zero, thus no real equal roots for any real $p$. **If question meant roots real (not necessarily equal), then:** $$D \geq 0 \Rightarrow (2p -4)^2 + 12 \geq 0$$ Always true for all real $p$. No $q$ given in the equation, so $q$ is not applicable here. 5. **Prove that: $$(\csc \theta - \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$$** Start with left side: $$\csc \theta - \cot \theta = \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} = \frac{1 - \cos \theta}{\sin \theta}$$ Square it: $$\left(\frac{1 - \cos \theta}{\sin \theta}\right)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$$ Use identity $\sin^2 \theta = 1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$: $$\frac{(1 - \cos \theta)^2}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 - \cos \theta}{1 + \cos \theta}$$ This matches the right side, so proved. 6. **Find the vector $2\vec{a} - 3\vec{b}$ where $\vec{a} = 3\hat{i} - 2\hat{j} + 5\hat{k}$ and $\vec{b} = -2\hat{i} - \hat{j} + \hat{k}$** Calculate: $$2\vec{a} = 6\hat{i} - 4\hat{j} + 10\hat{k}$$ $$3\vec{b} = -6\hat{i} - 3\hat{j} + 3\hat{k}$$ Therefore: $$2\vec{a} - 3\vec{b} = (6 + 6)\hat{i} + (-4 + 3)\hat{j} + (10 - 3)\hat{k} = 12\hat{i} - \hat{j} + 7\hat{k}$$ 7. **(a) Show that points (-1, 2), (7, 5), and (2, -6) are vertices of a right triangle** Calculate lengths: $$AB = \sqrt{(7 + 1)^2 + (5 - 2)^2} = \sqrt{8^2 + 3^2} = \sqrt{64 + 9} = \sqrt{73}$$ $$BC = \sqrt{(2 - 7)^2 + (-6 -5)^2} = \sqrt{(-5)^2 + (-11)^2} = \sqrt{25 + 121} = \sqrt{146}$$ $$CA = \sqrt{(-1 - 2)^2 + (2 + 6)^2} = \sqrt{(-3)^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73}$$ Pairs are $AB = CA = \sqrt{73}$ Check Pythagoras: $$AB^2 + CA^2 = 73 + 73 = 146 = BC^2$$ So triangle is right angled at $A$ or $C$. 7. **(b) Find lengths of sides of triangle ABC with vertices (1,3), (5,11), (9,5)** Calculate sides: $$AB = \sqrt{(5-1)^2 + (11-3)^2} = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}$$ $$BC = \sqrt{(9-5)^2 + (5-11)^2} = \sqrt{4^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$$ $$CA = \sqrt{(9-1)^2 + (5-3)^2} = \sqrt{8^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17}$$ **Final answers:** - $\sin 105^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$ - Line equation: $y = -2x - 5$ - Constant term: 490992 - No real equal roots for the quadratic for any real $p$ - Identity proved true - Vector $2\vec{a} - 3\vec{b} = 12\hat{i} - \hat{j} + 7\hat{k}$ - Points form a right triangle - Side lengths of triangle ABC: $4\sqrt{5}, 2\sqrt{13}, 2\sqrt{17}$