CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 15

Answer: (d) sin 0°

Explanation:
1. tan 45° = 1, so the expression is (1 − 1)/(1 + 1) = 0.
2. sin 0° = 0, so the answer is sin 0°.

Teacher's Note:
1. Substitute tan 45° = 1 first.
2. tan 90° is not defined, so (a) can never be correct.

Answer: (d) \( \tan^2 A \)

Explanation:
1. 1 + tan²A = sec²A and 1 + cot²A = cosec²A.
2. So the expression = sec²A / cosec²A = sin²A / cos²A = tan²A.

Teacher's Note:
1. Use the two identities 1 + tan²A = sec²A and 1 + cot²A = cosec²A.
2. sec² over cosec² turns into sin² over cos².

Answer: (d) 1

Explanation:
1. sec 30° = 2/√3 and cosec 60° = 2/√3.
2. Their ratio is 1.

Teacher's Note:
1. sec 30° = cosec 60° because 30° and 60° are complementary.
2. sec(90° − θ) = cosec θ.

Answer: (c) \( \sqrt{2}\sin\theta \)

Explanation:
1. Given cos θ + sin θ = √2 cos θ, so sin θ = (√2 − 1) cos θ.
2. Then cos θ − sin θ = cos θ − (√2 − 1) cos θ = (2 − √2) cos θ = √2(√2 − 1) cos θ = \( \sqrt{2}\sin\theta \).

Teacher's Note:
1. Express sin θ in terms of cos θ and substitute.
2. Factor √2 out of (2 − √2).

Answer: (b) \( \frac{2}{3} \)

Explanation:
1. cos⁴θ − sin⁴θ = (cos²θ − sin²θ)(cos²θ + sin²θ) = cos²θ − sin²θ.
2. So cos²θ − sin²θ = 2/3. Also cos²θ − sin²θ = 1 − 2 sin²θ.
3. Hence 1 − 2 sin²θ = 2/3.

Teacher's Note:
1. Factorise a⁴ − b⁴ as (a² − b²)(a² + b²).
2. cos²θ = 1 − sin²θ converts the result.

Answer: (d) a

Explanation:
1. Take a right triangle with opposite side a² − 1 and hypotenuse a² + 1. The adjacent side is √((a²+1)² − (a²−1)²) = 2a.
2. sec θ + tan θ = (a² + 1)/(2a) + (a² − 1)/(2a) = 2a²/(2a) = a.

Teacher's Note:
1. Build the triangle from the given ratio and find the third side by Pythagoras.
2. Both ratios share the denominator 2a.

Answer: (c) \( \frac{41}{40} \)

Explanation:
1. Divide numerator and denominator by cos θ: (tan θ + 1)/(tan θ − 1) = 5/4.
2. 4 tan θ + 4 = 5 tan θ − 5, so tan θ = 9.
3. (tan²θ + 1)/(tan²θ − 1) = (81 + 1)/(81 − 1) = 82/80 = 41/40.

Teacher's Note:
1. Dividing by cos θ turns sin and cos terms into tan θ.
2. Find tan θ first, then substitute.

Answer: (a) \( \frac{7}{4} \)

Explanation:
1. sec²θ − tan²θ = 1 (identity).
2. sec⁴θ − tan⁴θ = (sec²θ − tan²θ)(sec²θ + tan²θ) = 1 × 7/4 = 7/4.

Teacher's Note:
1. Factorise the difference of fourth powers.
2. The identity sec²θ − tan²θ = 1 makes the first bracket 1.

Answer: (c) 1

Explanation:
1. The product is (sec θ)² − (tan θ)² because 1/cot θ = tan θ.
2. sec²θ − tan²θ = 1.

Teacher's Note:
1. Recognise (x + y)(x − y) = x² − y².
2. 1/cot θ = tan θ.

Answer: (c) \( \cot^2A-\cos^2A \)

Explanation:
1. cot²A cos²A = cot²A (1 − sin²A) = cot²A − cot²A sin²A.
2. cot²A sin²A = cos²A, so the result is cot²A − cos²A.

Teacher's Note:
1. Replace cos²A by 1 − sin²A.
2. cot A · sin A = cos A.

Answer: (d) 2

Explanation:
1. Square both sides: tan²θ + cot²θ + 2 tan θ cot θ = 4.
2. Since tan θ cot θ = 1, tan²θ + cot²θ = 4 − 2 = 2.

Teacher's Note:
1. Squaring a sum introduces the product term 2 tan θ cot θ.
2. tan θ × cot θ is always 1.

Answer: (b) \( \frac{3}{4} \)

Explanation:
1. sin A = √(1 − 16/25) = 3/5.
2. tan A = sin A / cos A = (3/5)/(4/5) = 3/4.

Teacher's Note:
1. 4, 3, 5 is the Pythagorean triple here.
2. tan A = opposite/adjacent = 3/4.

Answer: (c) \( \frac{\sqrt{b^2-a^2}}{b} \)

Explanation:
1. cos θ = √(1 − sin²θ) = √(1 − a²/b²) = \( \frac{\sqrt{b^2-a^2}}{b} \).

Teacher's Note:
1. Use sin²θ + cos²θ = 1 and take the positive root for an acute angle.
2. Combine 1 − a²/b² over the common denominator b².

Answer: (b) \( \frac{1}{\sqrt{3}} \)

Explanation:
1. tan M = KL / LM = 4/(4√3) = \( \frac{1}{\sqrt{3}} \).

Teacher's Note:
1. tan of an angle is opposite over adjacent.
2. For angle M the opposite side is KL.

Answer: (a) 30°

Explanation:
1. tan M = 1/√3 = tan 30°, so ∠M = 30°.

Teacher's Note:
1. Recall tan 30° = 1/√3.
2. Match the value with the standard table.

Answer: (c) 60°

Explanation:
1. tan K = LM / KL = 4√3/4 = √3 = tan 60°, so ∠K = 60°.
2. Check: ∠K + ∠M = 90°.

Teacher's Note:
1. The two acute angles of a right triangle add up to 90°.
2. tan 60° = √3.

Answer: (c) \( \tan^2 M \)

Explanation:
1. Using the identity sec²M − tan²M = 1, we get sec²M − 1 = tan²M.

Teacher's Note:
1. Rearrange sec²θ = 1 + tan²θ.
2. This holds for every acute angle.

Answer: (a) 0

Explanation:
1. tan 45° = 1, so the expression = (1 − 1)/(1 + 1) = 0.

Teacher's Note:
1. Substitute tan 45° = 1.
2. A zero numerator with a non-zero denominator gives 0.

Answer: (d) 6 m

Explanation:
1. AD = DC = 6 m.
2. BD² = AB² − AD² = 72 − 36 = 36, so BD = 6 m.

Teacher's Note:
1. In an isosceles triangle, the line from the apex to the mid-point of the base is perpendicular to the base.
2. Use Pythagoras in triangle ABD.

Answer: (c) 45°

Explanation:
1. In triangle ABD, sin A = BD/AB = 6/(6√2) = 1/√2 = sin 45°, so ∠A = 45°.

Teacher's Note:
1. Use sine since BD and AB are known.
2. sin 45° = 1/√2.

Answer: (c) 45°

Explanation:
1. In triangle BDC, tan C = BD/DC = 6/6 = 1 = tan 45°, so ∠C = 45°.

Teacher's Note:
1. AB = BC, so the base angles A and C are equal.
2. tan 45° = 1.

Answer: (d) \( \sqrt{2} \)

Explanation:
1. sin A = sin 45° = 1/√2 and cos C = cos 45° = 1/√2.
2. Their sum is 2/√2 = \( \sqrt{2} \).

Teacher's Note:
1. Add the two equal values 1/√2.
2. 2/√2 = √2.

Answer: (c) 2

Explanation:
1. tan C = tan A = tan 45° = 1.
2. So tan²C + tan²A = 1 + 1 = 2.

Teacher's Note:
1. Both base angles are 45°.
2. tan 45° = 1.

Answer: (a) 1

Explanation:
1. sin 30° + cos 60° = 1/2 + 1/2 = 1.

Teacher's Note:
1. Read sin 30° and cos 60° from the table; both are 1/2.
2. This reflects sin θ = cos(90° − θ).

Answer: (b) −1

Explanation:
1. Numerator: (1/2)² − 0 = 1/4.
2. Denominator: 0 − (1/2)² = −1/4.
3. The ratio is −1.

Teacher's Note:
1. Square the values before subtracting.
2. cos 90° = 0 and sin 0° = 0.

Answer: (c) \( \frac{5}{4} \)

Explanation:
1. sin²90° + cos²60° = 1² + (1/2)² = 1 + 1/4 = 5/4.

Teacher's Note:
1. sin 90° = 1 and cos 60° = 1/2.
2. Add the squares after squaring each value.

Answer: (a) 1

Explanation:
1. sin²60° + cos²60° = 3/4 + 1/4 = 1.
2. This is the identity sin²θ + cos²θ = 1.

Teacher's Note:
1. Both terms use the same angle, so the identity applies directly.
2. Check with the values 3/4 and 1/4.

Answer: (d) \( \frac{13}{5} \)

Explanation:
1. cosec B = hypotenuse / side opposite B = AB / AC = 13/5.

Teacher's Note:
1. In the right triangle at C, AB is the hypotenuse and AC is opposite to angle B.
2. cosec is the reciprocal of sin.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The Reason is the standard identity, so it is true.
2. Applying it: tan θ = (√3/2)/(1/2) = √3, so the Assertion is true and follows from the Reason.

Teacher's Note:
1. Divide sin θ by cos θ to get tan θ.
2. Both given values are those of 60°, so tan 60° = √3 agrees.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The Reason is Pythagoras' theorem, so it is true.
2. With perpendicular 3 and base 4, the hypotenuse is √(9 + 16) = 5, so the Assertion is true and follows from the Reason.

Teacher's Note:
1. The hypotenuse is always the greatest side of a right triangle.
2. tan θ = 3/4 means a 3, 4, 5 triangle.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The Reason is the basic identity, so it is true.
2. x + y = 2 sin²θ + 2 cos²θ + 1 = 2(1) + 1 = 3, so the Assertion is true and follows from the Reason.

Teacher's Note:
1. Group the sin² and cos² terms with a common factor 2.
2. The identity sin²θ + cos²θ = 1 holds for every angle.

Answer: (d) A is false, but R is true.

Explanation:
1. sin A is a single symbol for the sine ratio of angle A, not a product of sin and A, so the Assertion is false.
2. The sine of an acute angle does increase as the angle increases, so the Reason is true.

Teacher's Note:
1. sin, cos and tan are names of ratios, not quantities to be multiplied by the angle.
2. Sine rises from 0 at 0° to 1 at 90°.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. tan A = 1 gives A = 45°, so 2 sin A cos A = 2 × (1/√2) × (1/√2) = 1. The Assertion is true.
2. The Reason is a true statement but has nothing to do with the Assertion, so it is not the explanation.

Teacher's Note:
1. tan A = 1 means sin A = cos A, so A = 45°.
2. A true Reason must actually support the Assertion to be its explanation.