1. Problem: Given $z = 3 + 4i$, find the modulus $|z|$ and the argument of $z$. Step 1: The modulus of a complex number $z = x + yi$ is given by $$|z| = \sqrt{x^2 + y^2}.$$ Here, $x=3$ and $y=4$. Step 2: Calculate modulus: $$|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$$ Step 3: The argument $\theta$ is given by $$\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{4}{3}\right).$$ Step 4: Calculate argument: $$\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 0.93 \text{ radians}.$$ 2. Problem: Solve the system using determinants: $$2x + 3y = 5,$$ $$3x - y = 4.$$ Step 1: Write coefficient matrix $A$ and constants vector $B$: $$A = \begin{bmatrix}2 & 3 \\ 3 & -1\end{bmatrix}, \quad B = \begin{bmatrix}5 \\ 4\end{bmatrix}.$$ Step 2: Calculate determinant of $A$: $$\det(A) = (2)(-1) - (3)(3) = -2 - 9 = -11.$$ Step 3: Calculate determinants for $x$ and $y$: $$\det(A_x) = \begin{vmatrix}5 & 3 \\ 4 & -1\end{vmatrix} = (5)(-1) - (3)(4) = -5 - 12 = -17,$$ $$\det(A_y) = \begin{vmatrix}2 & 5 \\ 3 & 4\end{vmatrix} = (2)(4) - (5)(3) = 8 - 15 = -7.$$ Step 4: Solve for $x$ and $y$: $$x = \frac{\det(A_x)}{\det(A)} = \frac{-17}{-11} = \frac{17}{11},$$ $$y = \frac{\det(A_y)}{\det(A)} = \frac{-7}{-11} = \frac{7}{11}.$$ 3. Problem: Prove $$(1 + \tan^2 A)(1 - \sin A) = 1 + \sin A.$$ Step 1: Recall identity: $$1 + \tan^2 A = \sec^2 A = \frac{1}{\cos^2 A}.$$ Step 2: Substitute: $$(1 + \tan^2 A)(1 - \sin A) = \frac{1}{\cos^2 A}(1 - \sin A).$$ Step 3: Multiply numerator and denominator by $(1 + \sin A)$: $$= \frac{(1 - \sin A)(1 + \sin A)}{\cos^2 A (1 + \sin A)} = \frac{1 - \sin^2 A}{\cos^2 A (1 + \sin A)}.$$ Step 4: Use Pythagorean identity $1 - \sin^2 A = \cos^2 A$: $$= \frac{\cos^2 A}{\cos^2 A (1 + \sin A)} = \frac{1}{1 + \sin A}.$$ Step 5: There is a mistake; re-examine step 3. Actually, multiply numerator and denominator by $(1 + \sin A)$ to rationalize denominator is not needed here. Instead, rewrite: Step 3 revised: Multiply out numerator: $$(1 + \tan^2 A)(1 - \sin A) = \frac{1 - \sin A}{\cos^2 A}.$$ Step 4: Express numerator and denominator in terms of sine and cosine: $$= \frac{1 - \sin A}{\cos^2 A} = \frac{(1 - \sin A)(1 + \sin A)}{\cos^2 A (1 + \sin A)} = \frac{1 - \sin^2 A}{\cos^2 A (1 + \sin A)}.$$ Step 5: Use $1 - \sin^2 A = \cos^2 A$: $$= \frac{\cos^2 A}{\cos^2 A (1 + \sin A)} = \frac{1}{1 + \sin A}.$$ Step 6: The original statement is $$(1 + \tan^2 A)(1 - \sin A) = 1 + \sin A,$$ but we found it equals $\frac{1}{1 + \sin A}$. So the original statement is incorrect as given. Possibly a typo; if the right side is $\frac{1}{1 + \sin A}$, the identity holds. 4. Problem: Find the 6th term and sum of first 6 terms of A.P. with $a=2$, $d=3$. Step 1: The $n$th term of A.P. is $$a_n = a + (n-1)d.$$ Step 2: Calculate 6th term: $$a_6 = 2 + (6-1) \times 3 = 2 + 15 = 17.$$ Step 3: Sum of first $n$ terms: $$S_n = \frac{n}{2} [2a + (n-1)d].$$ Step 4: Calculate sum of first 6 terms: $$S_6 = \frac{6}{2} [2 \times 2 + (6-1) \times 3] = 3 [4 + 15] = 3 \times 19 = 57.$$ 5. Problem: Find the coefficient of $x^4$ in $(2 - 3x)^6$. Step 1: Use binomial theorem: $$(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k.$$ Step 2: Here, $a=2$, $b=-3x$, $n=6$. The term with $x^4$ corresponds to $k=4$: $$T_{k+1} = \binom{6}{4} 2^{6-4} (-3x)^4.$$ Step 3: Calculate coefficient: $$\binom{6}{4} = 15,$$ $$2^{2} = 4,$$ $$(-3)^4 = 81.$$ Step 4: Coefficient of $x^4$ is: $$15 \times 4 \times 81 = 4860.$$ 6. Problem: Find $\det(A)$ for $$A = \begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0\end{bmatrix}.$$ Step 1: Use cofactor expansion along first row: $$\det(A) = 1 \times \begin{vmatrix}1 & 4 \\ 6 & 0\end{vmatrix} - 2 \times \begin{vmatrix}0 & 4 \\ 5 & 0\end{vmatrix} + 3 \times \begin{vmatrix}0 & 1 \\ 5 & 6\end{vmatrix}.$$ Step 2: Calculate minors: $$M_{11} = (1)(0) - (4)(6) = 0 - 24 = -24,$$ $$M_{12} = (0)(0) - (4)(5) = 0 - 20 = -20,$$ $$M_{13} = (0)(6) - (1)(5) = 0 - 5 = -5.$$ Step 3: Calculate determinant: $$\det(A) = 1 \times (-24) - 2 \times (-20) + 3 \times (-5) = -24 + 40 - 15 = 1.$$ 7. Problem: Expand $(1 - 2x)^3$ up to $x^3$. Step 1: Use binomial theorem: $$(1 - 2x)^3 = \sum_{k=0}^3 \binom{3}{k} 1^{3-k} (-2x)^k.$$ Step 2: Calculate terms: - $k=0$: $\binom{3}{0} 1^3 (-2x)^0 = 1$ - $k=1$: $\binom{3}{1} 1^2 (-2x)^1 = 3 \times (-2x) = -6x$ - $k=2$: $\binom{3}{2} 1^1 (-2x)^2 = 3 \times 4x^2 = 12x^2$ - $k=3$: $\binom{3}{3} 1^0 (-2x)^3 = 1 \times (-8x^3) = -8x^3$ Step 3: Write expansion: $$(1 - 2x)^3 = 1 - 6x + 12x^2 - 8x^3.$$ 8. Problem: Find angle between vectors $\mathbf{a} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}$ and $\mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}$. Step 1: Use formula: $$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}.$$ Step 2: Calculate dot product: $$\mathbf{a} \cdot \mathbf{b} = (2)(1) + (1)(-1) + (-1)(1) = 2 - 1 - 1 = 0.$$ Step 3: Since dot product is zero, vectors are perpendicular. Step 4: Therefore, $$\theta = 90^\circ.$$ 9. Problem: Solve for $\theta$: $$2 \sin^2 \theta - 3 \sin \theta + 1 = 0.$$ Step 1: Let $x = \sin \theta$, rewrite: $$2x^2 - 3x + 1 = 0.$$ Step 2: Solve quadratic: $$x = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}.$$ Step 3: Roots: $$x_1 = 1, \quad x_2 = \frac{1}{2}.$$ Step 4: Find $\theta$: - For $\sin \theta = 1$, $\theta = 90^\circ$ or $\frac{\pi}{2}$ radians. - For $\sin \theta = \frac{1}{2}$, $\theta = 30^\circ$ or $\frac{\pi}{6}$ radians, and also $150^\circ$ or $\frac{5\pi}{6}$ radians in $[0, 2\pi)$. 10. Problem: In G.P., $a=81$, $r=\frac{1}{3}$, find sum to infinity. Step 1: Sum to infinity formula for $|r|<1$: $$S_\infty = \frac{a}{1-r}.$$ Step 2: Calculate: $$S_\infty = \frac{81}{1 - \frac{1}{3}} = \frac{81}{\frac{2}{3}} = 81 \times \frac{3}{2} = 121.5.$$ 11. Problem: If $a,b,c$ are in A.P., show that $$b = \frac{a + c}{2}.$$ Step 1: By definition of A.P., the difference between consecutive terms is constant: $$b - a = c - b.$$ Step 2: Rearrange: $$2b = a + c.$$ Step 3: Divide both sides by 2: $$b = \frac{a + c}{2}.$$ 12. Problem: Find area of triangle with sides 7, 8, 9 using Heron's formula. Step 1: Calculate semi-perimeter: $$s = \frac{7 + 8 + 9}{2} = \frac{24}{2} = 12.$$ Step 2: Use Heron's formula: $$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12(12-7)(12-8)(12-9)}.$$ Step 3: Calculate: $$= \sqrt{12 \times 5 \times 4 \times 3} = \sqrt{720} = 12 \sqrt{5}.$$ Step 4: Approximate: $$12 \sqrt{5} \approx 12 \times 2.236 = 26.83.$$ Final answers: 1. $|z|=5$, argument $\approx 0.93$ radians. 2. $x=\frac{17}{11}$, $y=\frac{7}{11}$. 3. Identity as given is incorrect; corrected form is $(1 + \tan^2 A)(1 - \sin A) = \frac{1}{1 + \sin A}$. 4. 6th term = 17, sum of first 6 terms = 57. 5. Coefficient of $x^4$ is 4860. 6. $\det(A) = 1$. 7. Expansion: $1 - 6x + 12x^2 - 8x^3$. 8. Angle between vectors = $90^\circ$. 9. $\theta = 90^\circ, 30^\circ, 150^\circ$. 10. Sum to infinity = 121.5. 11. $b = \frac{a + c}{2}$. 12. Area $= 12 \sqrt{5} \approx 26.83$.