Trigonometry - 30 Questions CBSE Class X
Trigonometry Mastery
CBSE Class X | Q31-Q60 Advanced Problems
🏆 Samadhan Academy 2026 Revision
Hard Questions
Marks Each
Attempted
Completion
Overall Progress
Difficulty: ⭐⭐⭐
3 MarksProve that:
sin⁴θ + cos⁴θ = 1 - 2sin²θcos²θ
यदि sin⁴θ + cos⁴θ = 1 - 2sin²θcos²θ को सिद्ध करें
✓ Complete Proof:
LHS = sin⁴θ + cos⁴θ
= (sin²θ)² + (cos²θ)²
Using a² + b² = (a+b)² - 2ab:
= (sin²θ + cos²θ)² - 2sin²θcos²θ
= (1)² - 2sin²θcos²θ
= 1 - 2sin²θcos²θ = RHS ✓
Key Concept: Algebraic identities combined with fundamental trig identity sin²θ + cos²θ = 1
Difficulty: ⭐⭐⭐
3 MarksGiven Condition:
If x sin³θ + y cos³θ = sin θ cos θ
and x sin θ = y cos θ
Prove: x² + y² = 1
✓ Step-by-Step Solution:
Given: x sin θ = y cos θ
⟹ x/y = cos θ/sin θ
From first equation:
x sin³θ + y cos³θ = sin θ cos θ
Substitute y cos θ = x sin θ:
x sin³θ + (x sin θ/cos θ) × cos³θ = sin θ cos θ
x sin³θ + x sin θ cos²θ = sin θ cos θ
x sin θ(sin²θ + cos²θ) = sin θ cos θ
x sin θ = sin θ cos θ ⟹ x = cos θ
y cos θ = x sin θ ⟹ y = sin θ
x² + y² = cos²θ + sin²θ = 1 ✓
Difficulty: ⭐⭐⭐
3 MarksFind the value:
If sin θ + cos θ = √2
Find: sin⁶θ + cos⁶θ
जहां 0° < θ < 90°
✓ Detailed Solution:
Given: sin θ + cos θ = √2
Square both sides:
(sin θ + cos θ)² = 2
sin²θ + 2sin θ cos θ + cos²θ = 2
1 + 2sin θ cos θ = 2
sin θ cos θ = 1/2
Now, sin³θ + cos³θ = (sin θ + cos θ)³ - 3sin θ cos θ(sin θ + cos θ)
= (√2)³ - 3(1/2)(√2)
= 2√2 - 3√2/2 = √2/2
sin⁶θ + cos⁶θ = (sin²θ + cos²θ)³ - 3sin²θcos²θ(sin²θ + cos²θ) + 3sin⁴θcos⁴θ
sin⁶θ + cos⁶θ = 1/8
Difficulty: ⭐⭐⭐⭐
4 MarksProve that:
(tan θ + cot θ)² = sec²θ + cosec²θ
✓ Complete Proof:
LHS = (tan θ + cot θ)²
= tan²θ + 2tan θ cot θ + cot²θ
= tan²θ + 2(1) + cot²θ
= tan²θ + cot²θ + 2
= (sec²θ - 1) + (cosec²θ - 1) + 2
= sec²θ + cosec²θ = RHS ✓
Note: Key insight: tan θ · cot θ = 1, and use of derived identities
Difficulty: ⭐⭐⭐
3 MarksSolve for θ:
2sin²θ - 3sin θ + 1 = 0
(where 0° ≤ θ ≤ 90°)
✓ Solution:
Let sin θ = x
Equation: 2x² - 3x + 1 = 0
Factorize: 2x² - 2x - x + 1 = 0
2x(x - 1) - 1(x - 1) = 0
(2x - 1)(x - 1) = 0
x = 1/2 or x = 1
sin θ = 1/2 ⟹ θ = 30°
sin θ = 1 ⟹ θ = 90°
Difficulty: ⭐⭐⭐
3 MarksGiven Condition:
If 5sin θ = 3
Find: (sin θ - 2cos³θ) / (2sin³θ - cos θ)
✓ Solution:
Given: 5sin θ = 3 ⟹ sin θ = 3/5
cos²θ = 1 - sin²θ = 1 - 9/25 = 16/25
cos θ = 4/5
Divide numerator and denominator by cos³θ:
= (tan θ - 2) / (2tan³θ - 1)
tan θ = 3/4
= (3/4 - 2) / (2(27/64) - 1)
= (-5/4) / (27/32 - 1)
= (-5/4) / (-5/32)
= 8
Difficulty: ⭐⭐⭐⭐
4 MarksProve that:
√[(1 + sin θ)/(1 - sin θ)] = sec θ + tan θ
✓ Complete Proof:
LHS = √[(1 + sin θ)/(1 - sin θ)]
Rationalize: multiply by (1 + sin θ)/(1 + sin θ)
= √[(1 + sin θ)²/((1 - sin θ)(1 + sin θ))]
= √[(1 + sin θ)²/(1 - sin²θ)]
= √[(1 + sin θ)²/cos²θ]
= (1 + sin θ)/cos θ
= 1/cos θ + sin θ/cos θ
= sec θ + tan θ = RHS ✓
Difficulty: ⭐⭐⭐
3 MarksFind the value:
sin²θ + sin⁴θ = 1
Find: cos²θ + cos⁴θ
✓ Solution:
Given: sin²θ + sin⁴θ = 1
sin²θ(1 + sin²θ) = 1
Let sin²θ = x, then: x(1 + x) = 1
x + x² = 1 ⟹ x² + x - 1 = 0
x = (-1 ± √5)/2
Since x > 0: sin²θ = (√5 - 1)/2
cos²θ = (3 - √5)/2
cos⁴θ = [(3 - √5)/2]² = (14 - 6√5)/4 = (7 - 3√5)/2
cos²θ + cos⁴θ = (3 - √5)/2 + (7 - 3√5)/2 = (10 - 4√5)/2 = 5 - 2√5
Difficulty: ⭐⭐⭐⭐
4 MarksGiven Conditions:
tan(A + B) = √3
tan(A - B) = 1/√3
Find: A and B (0° < A, B < 90°)
✓ Solution:
tan(A + B) = √3 ⟹ A + B = 60°
tan(A - B) = 1/√3 ⟹ A - B = 30°
Add equations: 2A = 90° ⟹ A = 45°
Subtract: 2B = 30° ⟹ B = 15°
Difficulty: ⭐⭐⭐
3 MarksProve that:
(cosec θ + sec θ)(cos θ - sin θ) = cot θ - tan θ
✓ Complete Proof:
LHS = (cosec θ + sec θ)(cos θ - sin θ)
= (1/sin θ + 1/cos θ)(cos θ - sin θ)
= [(cos θ + sin θ)/(sin θ cos θ)](cos θ - sin θ)
= (cos²θ - sin²θ)/(sin θ cos θ)
= cos²θ/(sin θ cos θ) - sin²θ/(sin θ cos θ)
= cos θ/sin θ - sin θ/cos θ
= cot θ - tan θ = RHS ✓
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