Mathematics Section C Study Notes: Rationality, Geometry, and Trigonometry

Problem Statement: Prove that $\sqrt{2}$ is an irrational number.
Proof by Contradiction Methodology:
- Assume $\sqrt{2}$ is a rational number. By definition, it can be written in the form $\frac{p}{q}$ , where $p$ and $q$ are integers, $q \neq 0$ , and $p, q$ are co-prime (having no common factors other than 1).
- Equation: $\sqrt{2} = \frac{p}{q}$ .
- Squaring both sides: $2 = \frac{p^2}{q^2}$ , which implies $p^2 = 2q^2$ .
- Since $2$ divides $p^2$ , by number theory, $2$ must also divide $p$ . Thus, let $p = 2k$ for some integer $k$ .
- Substituting back: $(2k)^2 = 2q^2 \implies 4k^2 = 2q^2 \implies q^2 = 2k^2$ .
- This implies $2$ divides $q^2$ , and therefore $2$ must divide $q$ .
- Conclusion: Since both $p$ and $q$ share a common factor of $2$ , the initial assumption that they are co-prime is contradicted. Therefore, $\sqrt{2}$ must be irrational.

Part (a): Tangent Angle Relationship
- Theorem to Prove: Two tangents $PA$ and $PB$ are drawn to a circle with centre $O$ from an external point $P$ . Prove that $\angle APB = 2\angle OAB$ .
- Key Geometric Principles:
  - Tangents from an external point to a circle are equal in length ( $PA = PB$ ).
  - Triangle $PAB$ is an isosceles triangle ( $PA = PB$ ).
  - The angle between the radius and the tangent at the point of contact is $90^{\circ}$ ( $\angle OAP = 90^{\circ}$ ).
Part (b) OR Option: Calculating Chord and Tangent Properties
- Given Data:
  - $PA$ is a tangent to the circle with centre $O$ .
  - Radius segment $OA = 10\,cm$ .
  - Segment $AB = 8\,cm$ .
  - Perpendicular condition: $AB \perp OP$ .
- Goal: Find the length of $PB$ .

Problem Statement: Determine the ratio in which the line $3x + y - 9 = 0$ divides the line segment joining the points $(1, 3)$ and $(2, 5)$ . Find the point of intersection.
Calculation Steps:
- Section Formula: Let the ratio be $k:1$ . The coordinates of the point of intersection $(x, y)$ are:
  - $x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{2k + 1}{k + 1}$
  - $y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{5k + 3}{k + 1}$
- Equation Substitution: Since this point lies on the line $3x + y - 9 = 0$ , substitute the expressions for $x$ and $y$ into the equation to solve for $k$ .
- Point of Intersection: Once $k$ is found, substitute it back into the section formula to find the numerical $(x, y)$ coordinates.

Part (a): Condition-Based Proof
- Given: $\sin(\theta) + \cos(\theta) = \sqrt{3}$ .
- Goal: Prove that $\tan(\theta) + \cot(\theta) = 1$ .
- Step-by-Step Logic:
  1. Square the given equation: $(\sin(\theta) + \cos(\theta))^2 = (\sqrt{3})^2$ .
  2. Expand: $\sin^2(\theta) + \cos^2(\theta) + 2\sin(\theta)\cos(\theta) = 3$ .
  3. Using $\sin^2(\theta) + \cos^2(\theta) = 1$ , we get: $1 + 2\sin(\theta)\cos(\theta) = 3 \implies 2\sin(\theta)\cos(\theta) = 2 \implies \sin(\theta)\cos(\theta) = 1$ .
  4. Expand the expression to prove: $\tan(\theta) + \cot(\theta) = \frac{\sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{\sin(\theta)} = \frac{\sin^2(\theta) + \cos^2(\theta)}{\sin(\theta)\cos(\theta)}$ .
  5. Substitute known values: $\frac{1}{1} = 1$ . Proof complete.
Part (b) OR Option: Complex Identity Verification
- Identity to Prove: $(\sin(A) + \sec(A))^2 + (\cos(A) + \text{cosec}(A))^2 = (1 + \sec(A)\text{cosec}(A))^2$

Sub-Question (b): Square Coordinates
- (i) Find the coordinates of point $P$ .
- (ii) Find the length of the side of the square.
Alternative OR Option (Equidistant Point):
- Problem: Find the coordinates of a point on the line $x + y = 5$ which is equidistant from the points $(6, 4)$ and $(5, 2)$ .
- Mathematical Representation: Let the point be $Q(x, y)$ .
  1. Constraint 1: $x + y = 5$ , so $y = 5 - x$ .
  2. Constraint 2: Distance from $(6, 4)$ equals distance from $(5, 2)$ .
  3. Distance Formula: $\sqrt{(x-6)^2 + (y-4)^2} = \sqrt{(x-5)^2 + (y-2)^2}$ .
  4. Solve for $x$ and then solve for $y$ .