CBSE Class 10 Mathematics Real Numbers, Polynomials and Trigonometry Assignment Set 11

Answer: (c) 19. 95 = 5 × 19 and 152 = 8 × 19, so the HCF is 19.

Teacher's Note:
1. Write 95 = 5 × 19 and 152 = 8 × 19.
2. The common prime factor is 19.

Answer: (a) 1. 6ⁿ ends in 6 and 5ⁿ ends in 5, so their difference ends in 6 − 5 = 1.

Teacher's Note:
1. 6ⁿ always ends in 6 and 5ⁿ always ends in 5.
2. So the difference always ends in 1.

Answer: (d) equal. HCF ≤ each number ≤ LCM, so HCF = LCM only if the numbers are equal.

Teacher's Note:
1. HCF divides both numbers and LCM is a multiple of both.
2. They can be equal only if the numbers themselves are equal.

Answer: (b) −1. 1/α + 1/β = (α + β)/(αβ) = 1/(−1) = −1.

Teacher's Note:
1. Use the sum of zeroes = −b/a and product of zeroes = c/a.
2. Here the sum is 1 and the product is −1.

Answer: (b) 5. Product of zeroes = k/5 = 1, so k = 5.

Teacher's Note:
1. The product of the zeroes of a quadratic is c/a.
2. For reciprocal zeroes the product is 1.

Answer: (c) 2. The expression = [(cos θ + sin θ + 1)/cos θ] × [(sin θ + cos θ − 1)/sin θ] = [(sin θ + cos θ)² − 1]/(sin θ cos θ) = 2.

Teacher's Note:
1. Convert everything into sin θ and cos θ.
2. The numerator becomes (sin θ + cos θ)² − 1 = 2 sin θ cos θ.

Answer: (d) cos A. (sec A + tan A)(1 − sin A) = [(1 + sin A)/cos A] × (1 − sin A) = (1 − sin²A)/cos A = cos A.

Teacher's Note:
1. Write sec A and tan A in terms of sin A and cos A.
2. Then use 1 − sin²A = cos²A.

Answer: 12576 = 4052 × 3 + 420
4052 = 420 × 9 + 272
420 = 272 × 1 + 148
272 = 148 × 1 + 124
148 = 124 × 1 + 24
124 = 24 × 5 + 4
24 = 4 × 6 + 0
The remainder is now zero, so the HCF of 4052 and 12576 is 4.

Teacher's Note:
1. The divisor at the stage where the remainder becomes zero is the HCF.
2. Always divide the larger number by the smaller number first.

Answer: cot A = 8/15, so in a right triangle the base = 8k and perpendicular = 15k.
Hypotenuse = √(8² + 15²) k = √289 k = 17k.
sin A = 15/17 and sec A = 17/8.

Teacher's Note:
1. Draw a right triangle and apply the Pythagoras theorem.
2. cot A is the ratio of the adjacent side to the opposite side.

Answer: LHS: cot A − cos A = cos A/sin A − cos A = cos A (1/sin A − 1) = cos A (cosec A − 1).
Similarly cot A + cos A = cos A (cosec A + 1).
So LHS = [cos A (cosec A − 1)] / [cos A (cosec A + 1)] = (cosec A − 1)/(cosec A + 1) = RHS. Hence proved.

Teacher's Note:
1. Take cos A common from the numerator and the denominator.
2. Then cancel the common factor.

Answer: Let a be any positive integer. By Euclid's lemma, a = 3q + r with r = 0, 1 or 2.
Case 1: a = 3q, so a² = 9q² = 3(3q²) = 3m, where m = 3q².
Case 2: a = 3q + 1, so a² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1.
Case 3: a = 3q + 2, so a² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1 = 3m + 1.
Hence the square of any positive integer is of the form 3m or 3m + 1.

Teacher's Note:
1. Use the remainders 0, 1 and 2 when dividing by 3.
2. In each case, take 3 out as a common factor.

Answer: Group the terms: f(x) = x²(x − 5) − 16(x − 5) = (x − 5)(x² − 16) = (x − 5)(x − 4)(x + 4).
The zeroes are 5, 4 and −4; the two equal in magnitude but opposite in sign are 4 and −4.

Teacher's Note:
1. Factorisation by grouping gives all the zeroes quickly.
2. The sum 4 + (−4) + 5 = 5 also agrees with the coefficient of x².

Answer: cosec A = 2, so sin A = 1/2, A = 30°.
sin²A = 1/4, cot²A = 3, tan²A = 1/3, cos²A = 3/4.
Numerator = 2 × 1/4 + 3 × 3 = 1/2 + 9 = 19/2.
Denominator = 4 × 1/3 − 3/4 = 16/12 − 9/12 = 7/12.
Value = (19/2) × (12/7) = 114/7.

Teacher's Note:
1. Substitute the values of the ratios for 30° carefully.
2. Simplify the numerator and denominator separately before dividing.

Answer: Use cosec²A − cot²A = 1.
1/(cosec A − cot A) = (cosec A + cot A)/(cosec²A − cot²A) = cosec A + cot A.
1/(cosec A + cot A) = (cosec A − cot A)/(cosec²A − cot²A) = cosec A − cot A.
LHS = cosec A + cot A − cosec A = cot A.
RHS = cosec A − (cosec A − cot A) = cot A.
LHS = RHS, hence proved.

Teacher's Note:
1. Rationalise each denominator using the identity cosec²A − cot²A = 1.
2. Note that 1/sin A = cosec A.

Answer: cos 58° = sin 32°, so cos 58°/sin 32° = 1.
sin 22° = cos 68°, so sin 22°/cos 68° = 1.
cos 38° cosec 52° = cos 38°/sin 52° = cos 38°/cos 38° = 1.
tan 18° tan 72° = 1 and tan 35° tan 55° = 1, so the denominator = tan 60° = √3.
Value = 1 + 1 − \( \dfrac{1}{\sqrt{3}} \) = 2 − \( \dfrac{1}{\sqrt{3}} \) = \( \dfrac{6-\sqrt{3}}{3} \).

Teacher's Note:
1. Use the complementary angle relations sin (90° − θ) = cos θ and tan (90° − θ) = cot θ.
2. Pairs of angles that add up to 90° simplify the expression.

Answer: Divide the numerator and the denominator of the LHS by cos θ.
LHS = (tan θ − 1 + sec θ)/(tan θ + 1 − sec θ).
Replace 1 in the numerator by sec²θ − tan²θ = (sec θ − tan θ)(sec θ + tan θ):
Numerator = (sec θ + tan θ) − (sec θ − tan θ)(sec θ + tan θ) = (sec θ + tan θ)(1 − sec θ + tan θ).
Denominator = 1 + tan θ − sec θ.
So LHS = sec θ + tan θ = (sec²θ − tan²θ)/(sec θ − tan θ) = 1/(sec θ − tan θ) = RHS. Hence proved.

Teacher's Note:
1. Dividing by cos θ converts everything into tan θ and sec θ.
2. The identity sec²θ − tan²θ = 1 is the key step.