CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 17

Answer: (a) 1

Explanation:
1. sin²51° = sin²(90° − 39°) = cos²39°.
2. So sin²39° + sin²51° = sin²39° + cos²39° = 1.

Teacher's Note:
1. Use sin(90° − θ) = cos θ for complementary angles.
2. 39° + 51° = 90°.

Answer: (c) 0

Explanation:
1. tan²45° = 1 and sin 50° = cos 40°, so sin²40° + sin²50° = sin²40° + cos²40° = 1.
2. The numerator = 1 − 1 = 0.
3. The denominator tan 10° tan 80° = tan 10° cot 10° = 1, so the value is 0 ÷ 1 = 0.

Teacher's Note:
1. Convert one angle of each complementary pair.
2. tan(90° − θ) = cot θ.

Answer: (d) not defined

Explanation:
1. tan 90° = \( \frac{\sin 90^\circ}{\cos 90^\circ} = \frac{1}{0} \), which is not defined.

Teacher's Note:
1. Division by zero is undefined.
2. tan θ increases without limit as θ approaches 90°.

Answer: (b) 0

Explanation:
1. sin 27° = cos 63° and cos 27° = sin 63°.
2. sin²63° + sin²27° = sin²63° + cos²63° = 1 and cos²63° + cos²27° = cos²63° + sin²63° = 1.
3. The fraction is 1 ÷ 1 = 1 and sin²90° = 1, so the value is 1 − 1 = 0.

Teacher's Note:
1. Use complementary angles: 63° + 27° = 90°.
2. sin 90° = 1.

Answer: (a) 0

Explanation:
1. cot 65° = tan 25°, because 65° = 90° − 25°.
2. Numerator: tan²25° − cot 65° tan 25° = tan²25° − tan 25° tan 25° = 0.
3. So the whole expression is 0 ÷ (sin²25° + sin²65°) = 0.

Teacher's Note:
1. Replace cot 65° by tan 25° first.
2. The denominator is 1, but the numerator already gives 0.

Answer: (c) 0

Explanation:
1. cosec(58° + θ) = sec(90° − 58° − θ) = sec(32° − θ), so the numerator is 0.
2. tan(45° + θ) = cot(90° − 45° − θ) = cot(45° − θ), so the denominator = tan 45° = 1.
3. The value is 0 ÷ 1 = 0.

Teacher's Note:
1. Convert a cosec angle to sec using the complement.
2. The numerator being 0 gives the answer directly.

Answer: (c) sin(90° − A) = cos A.

Explanation:
1. For an acute angle A, the complement of A is 90° − A, and sin(90° − A) = cos A. So (c) is true.
2. The reciprocal of sin A is cosec A, so (a) is false.
3. cos 90° = 0, so (b) is false.
4. cos(90° − A) = sin A, so (d) is false.

Teacher's Note:
1. sin and cos swap when the angle is replaced by its complement.
2. Reciprocal of sin is cosec, of cos is sec, of tan is cot.

Answer: (b) tan 70° tan 20° = 1.

Explanation:
1. tan 70° = tan(90° − 20°) = cot 20°, so tan 70° tan 20° = cot 20° tan 20° = 1. (b) is true.
2. The ratios are defined using a right triangle, so (a) is false.
3. sec A and cosec A are never between −1 and 1, so (c) is false.
4. sin A = BC/AB and sin B = AC/AB are equal only when BC = AC, so (d) is false.

Teacher's Note:
1. Complementary angles give tan A tan(90° − A) = 1.
2. sec and cosec always have a magnitude of at least 1.

Answer: (d) sin θ = \( \frac{5}{3} \) for some angle θ.

Explanation:
1. The sine of any angle lies between −1 and 1, while \( \frac{5}{3} \) is greater than 1. So (d) is false.
2. sin²A + cos²A = 1 is the basic identity, so (a) is true.
3. sin²A is just a short way of writing (sin A)², so (b) is true.
4. cot 0° = cos 0° ÷ sin 0° = 1 ÷ 0, which is not defined, so (c) is true.

Teacher's Note:
1. The greatest value of sin θ is 1.
2. A ratio with 0 in the denominator is undefined.

Answer: (b) The expression (cos²20° − sin²67°) is positive.

Explanation:
1. sin 67° = cos 23°, so cos²20° − sin²67° = cos²20° − cos²23°.
2. The cosine decreases as the angle increases, so cos 20° > cos 23° and the expression is positive. (b) is true.
3. cot 0° is not defined, so (a) is false.
4. sec A ≥ 1 for an acute angle, so (c) is false.
5. tan 60° = \( \sqrt{3} \) > 1, so (d) is false.

Teacher's Note:
1. Convert to the same ratio before comparing.
2. Cosine decreases and sine increases in the first quadrant.