Exercise 8.4 - Q5(i)
Textbook Exercise Problem
Prove the identity: (cosec θ - cot θ)² = (1 - cos θ) / (1 + cos θ).
Step-by-Step Verified Solution:
- Step 1: Write LHS in terms of sin θ and cos θ: cosec θ = 1/sin θ, cot θ = cos θ/sin θ.
- Step 2: LHS = [(1 - cos θ) / sin θ]² = (1 - cos θ)² / sin²θ.
- Step 3: Use identity sin²θ = 1 - cos²θ: (1 - cos θ)² / (1 - cos²θ).
- Step 4: Factorize denominator as (1 - cos θ)(1 + cos θ):
- Step 5: [(1 - cos θ)(1 - cos θ)] / [(1 - cos θ)(1 + cos θ)] = (1 - cos θ) / (1 + cos θ) = RHS.
Final Answer: LHS = RHS = (1 - cos θ) / (1 + cos θ) (Proved)
Key Concept / Exam Strategy: Expressing all trigonometric terms in sine and cosine is the most reliable path to proof.
Generated Exercise #1
Foundation
Question: Evaluate: 2 tan²45° + cos²30° - sin²60°.
Core Concept Tested: Evaluation of trigonometric values at standard angles
Mentor Hint: tan 45° = 1, cos 30° = √3/2, sin 60° = √3/2.
Step-by-Step Solution:
- Step 1: tan²45° = (1)² = 1.
- Step 2: cos²30° = (√3/2)² = 3/4.
- Step 3: sin²60° = (√3/2)² = 3/4.
- Step 4: 2(1) + 3/4 - 3/4 = 2.
Verified Answer: 2
Generated Exercise #2
Intermediate
Question: If sec θ + tan θ = p, prove that sin θ = (p² - 1) / (p² + 1).
Core Concept Tested: Algebraic reciprocal property of sec²θ - tan²θ = 1
Mentor Hint: sec θ - tan θ = 1/p. Add and subtract to isolate sec θ and tan θ.
Step-by-Step Solution:
- Step 1: sec θ + tan θ = p (Eq 1). Since sec²θ - tan²θ = 1, sec θ - tan θ = 1/p (Eq 2).
- Step 2: Adding gives 2 sec θ = p + 1/p = (p² + 1)/p => sec θ = (p² + 1)/(2p).
- Step 3: Subtracting gives 2 tan θ = p - 1/p = (p² - 1)/p => tan θ = (p² - 1)/(2p).
- Step 4: sin θ = tan θ / sec θ = [(p² - 1)/(2p)] / [(p² + 1)/(2p)] = (p² - 1) / (p² + 1).
Verified Answer: sin θ = (p² - 1) / (p² + 1) (Proved)
Generated Exercise #3
Intermediate
Question: A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground inclined at 60°. Find the length of the string, assuming there is no slack.
Core Concept Tested: Heights & Distances application of sin ratio
Mentor Hint: sin 60° = Perpendicular / Hypotenuse = 60 / L.
Step-by-Step Solution:
- Step 1: sin 60° = √3/2 = Height / String Length = 60 / L.
- Step 2: L = (60 × 2) / √3 = 120 / √3.
- Step 3: Rationalize: (120√3) / 3 = 40√3 meters (≈ 69.28 m).
Verified Answer: 40√3 meters (≈ 69.28 m)
Generated Exercise #4
JEE-NEET Foundation
Question: From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the total height of the tower.
Core Concept Tested: Multi-triangle elevation and depression geometry
Mentor Hint: Use 45° triangle to find horizontal distance between building and tower.
Step-by-Step Solution:
- Step 1: Let horizontal distance = d. In lower triangle: tan 45° = 7 / d => d = 7 m.
- Step 2: In upper triangle: Let extra height above building = h. tan 60° = h / d => √3 = h / 7 => h = 7√3 m.
- Step 3: Total height of tower = 7 + h = 7 + 7√3 = 7(1 + √3) meters (≈ 19.12 m).
Verified Answer: 7(1 + √3) m (≈ 19.12 m)
Generated Exercise #5
Advanced HOTS
Question: Prove that: (sin θ - 2 sin³θ) / (2 cos³θ - cos θ) = tan θ.
Core Concept Tested: Trigonometric factoring and multiple angle identity derivation
Mentor Hint: Factor sin θ from numerator and cos θ from denominator.
Step-by-Step Solution:
- Step 1: Numerator = sin θ (1 - 2 sin²θ).
- Step 2: Denominator = cos θ (2 cos²θ - 1).
- Step 3: Note that 1 - 2 sin²θ = (sin²θ + cos²θ) - 2 sin²θ = cos²θ - sin²θ.
- Step 4: Also 2 cos²θ - 1 = 2 cos²θ - (sin²θ + cos²θ) = cos²θ - sin²θ.
- Step 5: Since (1 - 2 sin²θ) = (2 cos²θ - 1), they cancel out: LHS = sin θ / cos θ = tan θ.
Verified Answer: LHS = tan θ = RHS (Proved)