CBSE Class 10 Mathematics Introduction to Trigonometry Worksheet Set 10

Answer: Since sin θ = cos(90° − θ), 3θ − 30° = 90° − θ, so θ = 30°. Hence tan θ = \( \dfrac{1}{\sqrt{3}} \).

Teacher's Note:
1. Convert the sine into the cosine of the complementary angle.
2. Equate the two angles and solve.

Answer: Expanding gives (1 + 2 sin θ cos θ) + (1 − 2 sin θ cos θ) = 2.

Teacher's Note:
1. The middle terms cancel after expansion.
2. Use sin²θ + cos²θ = 1 in both brackets.

Answer: cos²θ − sin²θ = (1 − tan²θ)/(1 + tan²θ) = (1 − 9/16)/(1 + 9/16) = \( \dfrac{7}{25} \).

Teacher's Note:
1. Divide numerator and denominator by cos²θ.
2. Substitute tan²θ = 9/16.

Answer: (sec θ + tan θ)(sec θ − tan θ) = sec²θ − tan²θ = 1, so sec θ − tan θ = \( \dfrac{1}{p} \).

Teacher's Note:
1. Use the identity sec²θ − tan²θ = 1.
2. Divide by the given expression.

Answer: sec⁴θ − sec²θ = sec²θ(sec²θ − 1) = (1 + tan²θ) tan²θ = tan²θ + tan⁴θ.

Teacher's Note:
1. Factor out sec²θ first.
2. Replace sec²θ with 1 + tan²θ.

Answer: LHS = sin θ(1 − 2 sin²θ) / [cos θ(2 cos²θ − 1)]. Since 2 cos²θ − 1 = 1 − 2 sin²θ, the common factor cancels, leaving sin θ / cos θ = tan θ.

Teacher's Note:
1. Factor sin θ from the numerator and cos θ from the denominator.
2. Show that the brackets are equal using sin²θ + cos²θ = 1.

Answer: tan A = 1/√3 gives A = 30°, so B = 60°. The value = sin 30° cos 60° + cos 30° sin 60° = ¼ + ¾ = 1.

Teacher's Note:
1. Find angle A from the given tangent value.
2. Use A + B = 90° because ∠C is a right angle.

Answer: tan 23° = cot 67°, so the expression = cosec²67° − cot²67° = 1.

Teacher's Note:
1. Convert tan 23° to a ratio of 67°.
2. Use cosec²θ − cot²θ = 1.

Answer: The right side is cos(60° − 30°) = cos 30°, so x = 30°.

Teacher's Note:
1. Recognise the formula cos(A − B).
2. Alternatively evaluate numerically: (1/2)(√3/2) + (√3/2)(1/2) = √3/2.

Answer: sin²x = 1/4, so sin x = 1/2 and x = 30°.

Teacher's Note:
1. Divide by 2 and take the square root.
2. Use the standard value of sin 30°.

Answer: cosec²30° = 4, sin²45° = 1/2, sec²60° = 4. So 4 − 1/2 − 4 = −\( \dfrac{1}{2} \).

Teacher's Note:
1. Substitute the standard values.
2. Do the subtraction carefully.

Answer: = (1 + sin θ)/cos θ × (1 − sin θ) = (1 − sin²θ)/cos θ = cos²θ/cos θ = cos θ.

Teacher's Note:
1. Write secant and tangent in terms of sine and cosine.
2. Use (1 + sin θ)(1 − sin θ) = cos²θ.

Answer: LHS = cos A [(1 + sin A) + (1 − sin A)] / (1 − sin²A) = 2 cos A / cos²A = 2 sec A.

Teacher's Note:
1. Take the common denominator (1 − sin A)(1 + sin A).
2. It equals cos²A.

Answer: tan²α = sec²α − 1 = 9/16, so tan α = 3/4. The value = (3/4)/(25/16) = \( \dfrac{12}{25} \).

Teacher's Note:
1. Find tan α using the identity.
2. Then use 1 + tan²α = sec²α = 25/16.

Answer: 3(sin²A + cos²A) + 4 sin²A = 4, so 4 sin²A = 1 and sin A = 1/2. Hence A = 30° and tan A = 1/√3.

Teacher's Note:
1. Write 7 sin²A as 3 sin²A + 4 sin²A.
2. Use sin²A + cos²A = 1.

Answer: LHS = sin²θ + cosec²θ + 2 + cos²θ + sec²θ + 2 = 1 + 4 + (1 + cot²θ) + (1 + tan²θ) = 7 + tan²θ + cot²θ.

Teacher's Note:
1. Expand each square; the cross terms give 2 each.
2. Use cosec²θ = 1 + cot²θ and sec²θ = 1 + tan²θ.

Answer: A − B = 60° and A + B = 90°. Adding gives A = 75° and B = 15°.

Teacher's Note:
1. Identify the angles whose tan is √3 and sin is 1.
2. Solve the two simultaneous equations.

Answer: sin θ = cos θ gives θ = 45°. The value = 3(1) + 2(1/2) − 1 = 3.

Teacher's Note:
1. Find θ from the given equality.
2. Substitute standard values for 45°.

Answer: LHS = (sec²θ − 1) + (cosec²θ − 1) + 2 = sec²θ + cosec²θ = (sin²θ + cos²θ)/(sin²θ cos²θ) = sec²θ cosec²θ.

Teacher's Note:
1. Convert tan²θ and cot²θ using the identities.
2. Combine the remaining two terms over a common denominator.

Answer: From the given equation sin θ = (√2 − 1) cos θ. Then cos θ − sin θ = (2 − √2) cos θ = √2(√2 − 1) cos θ = √2 sin θ.

Teacher's Note:
1. Express sin θ in terms of cos θ.
2. Substitute it into cos θ − sin θ.

Answer: sin⁶θ + cos⁶θ = (sin²θ + cos²θ)³ − 3 sin²θ cos²θ(sin²θ + cos²θ) = 1 − 3 sin²θ cos²θ.

Teacher's Note:
1. Use a³ + b³ = (a + b)³ − 3ab(a + b).
2. Put sin²θ + cos²θ = 1.

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

Teacher's Note:
1. Write every ratio in sine and cosine.
2. Use (a − 1)(a + 1) = a² − 1.

Answer: m² − 1 = 2 sin θ cos θ, and n = (sin θ + cos θ)/(sin θ cos θ) = m/(sin θ cos θ). So n(m² − 1) = m/(sin θ cos θ) × 2 sin θ cos θ = 2m.

Teacher's Note:
1. Find m² − 1 and n in terms of sin θ cos θ.
2. Multiply the two expressions.

Answer: Since sec θ − tan θ = 1/p, sec θ = (p + 1/p)/2 and tan θ = (p − 1/p)/2. So sin θ = tan θ/sec θ = (p² − 1)/(p² + 1).

Teacher's Note:
1. Use sec θ − tan θ = 1/p and solve for both ratios.
2. Divide tan θ by sec θ.

Answer: sin 65° = cos 25°, so the first two terms give 1. tan 5° tan 85° = 1 and tan 15° tan 75° = 1, tan 30° = 1/√3, so the bracket = 1/√3 and the product term is 1. Total = 2.

Teacher's Note:
1. Pair complementary angles.
2. Use tan θ tan(90° − θ) = 1.

Answer: Multiply numerator and denominator by (1 + cos A + sin A). Denominator becomes (1 + cos A)² − sin²A = 2 cos A(1 + cos A). Numerator becomes (1 + cos A)² + 2 sin A(1 + cos A) + sin²A = 2(1 + cos A)(1 + sin A). Dividing gives (1 + sin A)/cos A.

Teacher's Note:
1. Rationalise by multiplying with the numerator.
2. Then cancel the common factor (1 + cos A).