✅ STEP-BY-STEP PROOF (By contradiction method):

Step 1: Assume, to the contrary, that \(\sqrt{5}\) is rational.
Then \(\sqrt{5} = \frac{p}{q}\) where \(p\) and \(q\) are co-prime integers (no common factor other than 1) and \(q \neq 0\).

Step 2: Squaring both sides:

$$5 = \frac{p^2}{q^2}$$

$$p^2 = 5q^2 …(1)$$

Step 3: This implies that $$p^2$$ is divisible by 5.

If a prime number (5) divides $$p^2$$, then it also divides p.

So, p = 5k for some integer k.

Step 4: Substitute p = 5k in equation (1):

$$(5k)^2 = 5q^2$$

$$25k^2 = 5q^2$$

$$5k^2 = q^2$$

This implies $$q^2$$ is divisible by 5, so q is also divisible by 5.

Step 5: Thus, p and q both have common factor 5.

This contradicts our assumption that p and q are co-prime.

Step 6: Therefore, our assumption is false.

Hence, $$\sqrt{5}$$ is irrational.

✅ Hence proved: 

$${\sqrt{5} \text{ is irrational}}$$

✅ STEP-BY-STEP SOLUTION:

Step 1: Points of trisection divide the line segment in the ratio 1:2 and 2:1.
Let P and Q be the points of trisection with P closer to A and Q closer to B.

Step 2: For point P dividing AB in ratio 1:2 (AP:PB = 1:2)

Using section formula: P(x,y) $$= \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$$

Here m:n = 1:2, $$A(x_1,y_1) = (-1,4),$$

$$B(x_2,y_2) = (-3,-2)$$

$$x_P = \frac{1(-3) + 2(-1)}{1+2}$$

$$= \frac{-3 – 2}{3} = \frac{-5}{3}$$

$$y_P = \frac{1(-2) + 2(4)}{3}$$

$$= \frac{-2 + 8}{3} = \frac{6}{3} = 2$$

So, $$P = \left(-\frac{5}{3}, 2\right)$$

Step 3: For point Q dividing AB in ratio 2:1 (AQ:QB = 2:1)

$$x_Q = \frac{2(-3) + 1(-1)}{2+1}$$

$$= \frac{-6 – 1}{3} = \frac{-7}{3}$$

$$y_Q = \frac{2(-2) + 1(4)}{3}$$

$$= \frac{-4 + 4}{3} = \frac{0}{3} = 0$$

$$So, Q = \left(-\frac{7}{3}, 0\right)$$

✅ Final Answer: Points of trisection are $${\left(-\frac{5}{3}, 2\right)}$$ and $${\left(-\frac{7}{3}, 0\right)}$$

✅ STEP-BY-STEP PROOF:

Step 1: Note that \(\sec^2\theta – 1 = \tan^2\theta\) and \(\cos^2\theta – 1 = -\sin^2\theta\)

Step 2: First term: $$\frac{\sec^3\theta}{\sec^2\theta – 1}$$

$$= \frac{\sec^3\theta}{\tan^2\theta}$$

$$= \frac{\sec^3\theta}{\frac{\sin^2\theta}{\cos^2\theta}}$$

$$= \frac{\sec^3\theta \cdot \cos^2\theta}{\sin^2\theta}$$

$$= \frac{\frac{1}{\cos^3\theta} \cdot \cos^2\theta}{\sin^2\theta}$$

$$= \frac{1}{\cos\theta \cdot \sin^2\theta}$$

$$= \frac{\sec\theta}{\sin^2\theta}$$

Step 3: Second term: $$\frac{\cos^3\theta}{\cos^2\theta – 1}$$

$$= \frac{\cos^3\theta}{-\sin^2\theta} $$

$$= -\frac{\cos^3\theta}{\sin^2\theta}$$

Step 4: LHS

$$= \frac{\sec\theta}{\sin^2\theta} – \frac{\cos^3\theta}{\sin^2\theta}$$

$$= \frac{\sec\theta – \cos^3\theta}{\sin^2\theta}$$

Step 5: simplifying numerator from result of step 4

Write $$\sec\theta = \frac{1}{\cos\theta}$$

$$=\sec\theta – \cos^3\theta$$

$$= \frac{1}{\cos\theta} – \cos^3\theta = \frac{1 – \cos^4\theta}{\cos\theta} $$

$$= \frac{(1 – \cos^2\theta)(1 + \cos^2\theta)}{\cos\theta}$$

$$= \frac{\sin^2\theta(1 + \cos^2\theta)}{\cos\theta}$$

Step 6: Therefore, LHS$$ = \frac{\frac{\sin^2\theta(1 + \cos^2\theta)}{\cos\theta}}{\sin^2\theta}$$

$$= \frac{1 + \cos^2\theta}{\cos\theta}$$

$$= \frac{1}{\cos\theta} + \cos\theta = \sec\theta + \cos\theta$$

= RHS

✅ Hence proved

✅ STEP-BY-STEP PROOF:

Step 1: Given: \(p = \frac{\sec\alpha}{\cos\beta}\) and \(q = \frac{\tan\alpha}{\cos\beta}\)

Step 2: Find $$p^2 – q^2:$$

$$p^2 – q^2 $$

$$= \frac{\sec^2\alpha}{\cos^2\beta} – \frac{\tan^2\alpha}{\cos^2\beta} $$

$$= \frac{\sec^2\alpha – \tan^2\alpha}{\cos^2\beta}$$

Step 3: Using identity $$\sec^2\alpha – \tan^2\alpha = 1:$$

$$p^2 – q^2 = \frac{1}{\cos^2\beta}$$

Step 4: Multiply both sides by $$\sec^2\alpha = \frac{1}{\cos^2\alpha}$$

Step 3 become

$$=>(p^2 – q^2)\sec^2\alpha$$

$$= \frac{1}{\cos^2\beta} \cdot \frac{1}{\cos^2\alpha}$$

$$= \frac{1}{\cos^2\alpha \cos^2\beta}$$

Step 5: From given p $$= \frac{\sec\alpha}{\cos\beta} $$

Step 3 become

$$=> \frac{1}{\cos\alpha \cos\beta}$$

So,$$ p^2 $$

$$= \frac{1}{\cos^2\alpha \cos^2\beta}$$

Step 6: Therefore, $$(p^2 – q^2)\sec^2\alpha = p^2$$

✅ Hence proved

✅ STEP-BY-STEP PROOF:

Given: Circle with centre O. P is an external point. PA and PB are two tangents from P to the circle touching at A and B respectively.

To prove: PA = PB

Construction: Join OA, OB, and OP.

Proof:

Step 1: OA ⟂ PA and OB ⟂ PB (Radius is perpendicular to tangent at point of contact)

∴ ∠OAP = ∠OBP = 90°

Step 2: In right triangles OAP and OBP:

OA = OB (radii of same circle)

OP = OP (common side)

Step 3: By RHS (Right angle-Hypotenuse-Side) congruence criterion:

ΔOAP ≅ ΔOBP

Step 4: Corresponding parts of congruent triangles are equal:

PA = PB

✅ Hence proved: Lengths of tangents from an external point are equal.

✂️━━━━━━━━━━━━━━━━━━━

✅ STEP-BY-STEP PROOF:

Given: Circle with centre O. TP and TQ are tangents from external point T touching the circle at P and Q respectively.

To prove: ∠PTQ = 2∠OPQ

Construction: Join OP, OQ, and PQ.

Proof:

Step 1: OP ⟂ TP and OQ ⟂ TQ (Radius ⟂ tangent)

∴ ∠OPT = ∠OQT = 90°

Step 2: In quadrilateral OPTQ:

∠OPT + ∠OQT + ∠PTQ + ∠POQ = 360° (sum of angles of quadrilateral)

90° + 90° + ∠PTQ + ∠POQ = 360°

∠PTQ + ∠POQ = 180° …(1)

Step 3: In ΔOPQ, OP = OQ (radii)

∴ ∠OPQ = ∠OQP (angles opposite equal sides)

∠OPQ + ∠OQP + ∠POQ = 180°

2∠OPQ + ∠POQ = 180° …(2)

Step 4: From (1) and (2):

∠PTQ + ∠POQ = 2∠OPQ + ∠POQ

∠PTQ = 2∠OPQ

✅ Hence proved: \angle PTQ = 2\angle OPQ

✅ STEP-BY-STEP SOLUTION:

Step 1: Given: Radius \(r = 42\) cm, Central angle \(\theta = 30^\circ\), \(\pi = \frac{22}{7}\)

Step 2: Area of sector

$$= \frac{\theta}{360^\circ} \times \pi r^2$$

$$= \frac{30}{360} \times \frac{22}{7} \times (42)^2$$

$$= \frac{1}{12} \times \frac{22}{7} \times 1764$$

$$= \frac{1}{12} \times \frac{22}{7} \times 1764$$

Step 3: Simplify:

$$1764 \div 7 = 252$$

$$= \frac{1}{12} \times 22 \times 252$$

$$= \frac{1}{12} \times 5544$$

$$= 462 cm²$$

Step 4: Area of major sector = Area of circle – Area of minor sector

Area of circle

$$= \pi r^2 = \frac{22}{7} \times 1764 $$

$$= 22 \times 252 = 5544 cm²$$

Area of major sector = 5544 – 462 = 5082 cm²

✅ Final Answer:

Area of minor sector =$${462 \text{ cm}^2}$$

Area of major sector = $${5082 \text{ cm}^2}$$

✅ STEP-BY-STEP SOLUTION:

Step 1: Total outcomes when two dice are thrown = \(6 \times 6 = 36\)

Step 2 (i): Sum of numbers is 5

Favorable outcomes: (1,4), (2,3), (3,2), (4,1)

Number of favorable outcomes = 4

$$P(sum = 5) = \frac{4}{36} = \frac{1}{9}$$

Step 3 (ii): Difference of numbers is 3

Favorable outcomes: (1,4), (2,5), (3,6), (4,1), (5,2), (6,3)

Number of favorable outcomes = 6

$$P(difference = 3) = \frac{6}{36} = \frac{1}{6}$$

✅ Final Answer:

(i) P(sum = 5) = $${\frac{1}{9}}$$

(ii) P(difference = 3) =$$boxed{\frac{1}{6}}$$