Practice Questions For CBSE Class X
त्रिकोणमिति Trigonometry
Important Ratios & Identities | महत्वपूर्ण अनुपात और सर्वसमिकाएं
📚 Samadhan Academy 2026 Revision
📐 Trigonometric Ratios | त्रिकोणमितीय अनुपात
Based on Right-Angled Triangle | समकोण त्रिभुज पर आधारित
sin θ = p/h = लम्ब/कर्ण
cos θ = b/h = आधार/कर्ण
tan θ = p/b = लम्ब/आधार
cosec θ = h/p = 1/sin θ
sec θ = h/b = 1/cos θ
cot θ = b/p = 1/tan θ
💡 याद करने की ट्रिक | Memory Tricks
"पंडित बद्री प्रसाद"
P/H, B/H, P/B
sin, cos, tan के लिए
"Some People Have"
S=P/H (sin = perpendicular/hypotenuse)
"Curly Brown Hair" for cos, tan
🔢 Trigonometric Identities | त्रिकोणमितीय सर्वसमिकाएं
Must memorize for board exams | बोर्ड परीक्षा के लिए जरूरी
⭐ Fundamental Identities | मूल सर्वसमिकाएं
sin²θ + cos²θ = 1
पहली सर्वसमिका
1 + tan²θ = sec²θ
दूसरी सर्वसमिका
1 + cot²θ = cosec²θ
तीसरी सर्वसमिका
📝 Derived Forms | व्युत्पन्न रूप
From Identity 1:
sin²θ = 1 - cos²θ
cos²θ = 1 - sin²θ
From Identity 2:
tan²θ = sec²θ - 1
sec²θ - tan²θ = 1
From Identity 3:
cot²θ = cosec²θ - 1
cosec²θ - cot²θ = 1
Relationship:
tan θ = sin θ / cos θ
cot θ = cos θ / sin θ
🔄 Complementary Angles | पूरक कोण (90° - θ)
sin(90° - θ) = cos θ
cos(90° - θ) = sin θ
tan(90° - θ) = cot θ
cot(90° - θ) = tan θ
sec(90° - θ) = cosec θ
cosec(90° - θ) = sec θ
📊 Standard Values | मानक मान
Values of trigonometric ratios at standard angles
| Angle (θ) | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin θ | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan θ | 0 | 1/√3 | 1 | √3 | ∞ |
| cosec θ | ∞ | 2 | √2 | 2/√3 | 1 |
| sec θ | 1 | 2/√3 | √2 | 2 | ∞ |
| cot θ | ∞ | √3 | 1 | 1/√3 | 0 |
💡 याद करने का तरीका
sin θ: √0/2, √1/2, √2/2, √3/2, √4/2
= 0, 1/2, 1/√2, √3/2, 1
cos θ: Reverse of sin θ
= 1, √3/2, 1/√2, 1/2, 0
📝 Important Notes
- • sin और cos का मान 0 से 1 के बीच होता है
- • tan 90° = ∞ (अपरिभाषित / Not Defined)
- • cosec, sec, cot reciprocal हैं
- • 1/√2 = √2/2 ≈ 0.707
- • √3/2 ≈ 0.866
✏️ Practice Questions | अभ्यास प्रश्न
30 Important Questions for Board Exam Preparation
Questions Attempted
Click "Show Answer" to mark as attempted
If sin A = 3/5, find cos A.
यदि sin A = 3/5, तो cos A ज्ञात करें।
Answer: cos A = 4/5
Using sin²A + cos²A = 1
cos²A = 1 - (3/5)² = 1 - 9/25 = 16/25
cos A = 4/5
Find the value of sin 30° × cos 60°.
sin 30° × cos 60° का मान ज्ञात करें।
Answer: 1/4
sin 30° = 1/2, cos 60° = 1/2
sin 30° × cos 60° = 1/2 × 1/2 = 1/4
Find the value of tan 45° + cot 45°.
tan 45° + cot 45° का मान ज्ञात करें।
Answer: 2
tan 45° = 1, cot 45° = 1
tan 45° + cot 45° = 1 + 1 = 2
If tan θ = 4/3, find sin θ.
यदि tan θ = 4/3, तो sin θ ज्ञात करें।
Answer: sin θ = 4/5
tan θ = P/B = 4/3
H = √(P² + B²) = √(16 + 9) = 5
sin θ = P/H = 4/5
Find the value of sin²60° + cos²60°.
sin²60° + cos²60° का मान ज्ञात करें।
Answer: 1
Using identity: sin²θ + cos²θ = 1
This is true for any angle θ.
Find sec 0°.
sec 0° का मान ज्ञात करें।
Answer: 1
sec 0° = 1/cos 0° = 1/1 = 1
Find the value of sin(90° - 30°).
sin(90° - 30°) का मान ज्ञात करें।
Answer: √3/2
sin(90° - θ) = cos θ
sin(90° - 30°) = cos 30° = √3/2
If cos A = 1/2, find the value of A (0° ≤ A ≤ 90°).
यदि cos A = 1/2, तो A का मान ज्ञात करें।
Answer: A = 60°
cos 60° = 1/2
Therefore, A = 60°
Find the value of 2 sin 30° cos 30°.
2 sin 30° cos 30° का मान ज्ञात करें।
Answer: √3/2
2 sin 30° cos 30° = 2 × (1/2) × (√3/2) = √3/2
If sin θ = cos θ, find θ (0° < θ < 90°).
यदि sin θ = cos θ, तो θ ज्ञात करें।
Answer: θ = 45°
sin θ = cos θ only when θ = 45°
sin 45° = cos 45° = 1/√2
Prove that: (1 + tan²θ) cos²θ = 1
सिद्ध करें: (1 + tan²θ) cos²θ = 1
Proof:
LHS = (1 + tan²θ) cos²θ
= sec²θ × cos²θ (∵ 1 + tan²θ = sec²θ)
= (1/cos²θ) × cos²θ
= 1 = RHS ✓
Find the value of: (sin 45° + cos 45°)²
(sin 45° + cos 45°)² का मान ज्ञात करें।
Answer: 2
= (1/√2 + 1/√2)²
= (2/√2)²
= (√2)²
= 2
If tan A = √3 and tan B = 1/√3, find A + B.
यदि tan A = √3 और tan B = 1/√3, तो A + B ज्ञात करें।
Answer: 90°
tan A = √3 ⟹ A = 60°
tan B = 1/√3 ⟹ B = 30°
A + B = 60° + 30° = 90°
Evaluate: (sin²30° - cos²60°)/(tan²45° - cot²45°)
(sin²30° - cos²60°)/(tan²45° - cot²45°) का मान ज्ञात करें।
Answer: Not Defined (Undefined)
Numerator = (1/2)² - (1/2)² = 0
Denominator = 1² - 1² = 0
= 0/0 (Indeterminate form)
If sin(A + B) = 1 and cos(A - B) = 1, find A and B.
यदि sin(A + B) = 1 और cos(A - B) = 1, तो A और B ज्ञात करें।
Answer: A = 45°, B = 45°
sin(A + B) = 1 ⟹ A + B = 90°
cos(A - B) = 1 ⟹ A - B = 0°
Solving: A = 45°, B = 45°
Simplify: sin θ × cosec θ + cos θ × sec θ
सरल करें: sin θ × cosec θ + cos θ × sec θ
Answer: 2
= sin θ × (1/sin θ) + cos θ × (1/cos θ)
= 1 + 1
= 2
If 3 tan θ = 4, find (4 cos θ - sin θ)/(4 cos θ + sin θ).
यदि 3 tan θ = 4, तो (4 cos θ - sin θ)/(4 cos θ + sin θ) का मान ज्ञात करें।
Answer: 1/2
tan θ = 4/3
Divide numerator and denominator by cos θ:
= (4 - tan θ)/(4 + tan θ)
= (4 - 4/3)/(4 + 4/3)
= (8/3)/(16/3) = 1/2
Prove that: (1 - cos²θ)(1 + cot²θ) = 1
सिद्ध करें: (1 - cos²θ)(1 + cot²θ) = 1
Proof:
LHS = sin²θ × cosec²θ
= sin²θ × (1/sin²θ)
= 1 = RHS ✓
Find: tan 60° × cot 30° + sec²45°
tan 60° × cot 30° + sec²45° का मान ज्ञात करें।
Answer: 5
= √3 × √3 + (√2)²
= 3 + 2
= 5
If cot θ = 7/8, evaluate: (1 + sin θ)(1 - sin θ)/(1 + cos θ)(1 - cos θ)
यदि cot θ = 7/8, तो दिए गए व्यंजक का मान ज्ञात करें।
Answer: 49/64
= (1 - sin²θ)/(1 - cos²θ)
= cos²θ/sin²θ
= cot²θ
= (7/8)² = 49/64
Prove that: (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan²θ + cot²θ
सिद्ध करें।
Proof:
LHS = sin²θ + 2 + cosec²θ + cos²θ + 2 + sec²θ
= (sin²θ + cos²θ) + 4 + (1 + cot²θ) + (1 + tan²θ)
= 1 + 4 + 2 + tan²θ + cot²θ
= 7 + tan²θ + cot²θ = RHS ✓
Prove: (tan θ + sec θ - 1)/(tan θ - sec θ + 1) = (1 + sin θ)/cos θ
सिद्ध करें।
Proof:
Using sec²θ - tan²θ = 1:
LHS numerator = tan θ + sec θ - (sec²θ - tan²θ)
= (tan θ + sec θ)(1 - sec θ + tan θ) / denominator
After simplification = (1 + sin θ)/cos θ = RHS ✓
Prove: √[(1 - sin θ)/(1 + sin θ)] = sec θ - tan θ
सिद्ध करें।
Proof:
LHS = √[(1-sinθ)(1-sinθ)/(1+sinθ)(1-sinθ)]
= √[(1-sinθ)²/(1-sin²θ)]
= √[(1-sinθ)²/cos²θ]
= (1-sinθ)/cosθ
= 1/cosθ - sinθ/cosθ
= secθ - tanθ = RHS ✓
If tan θ + sin θ = m and tan θ - sin θ = n, prove that m² - n² = 4√(mn)
यदि tan θ + sin θ = m और tan θ - sin θ = n, तो सिद्ध करें।
Proof:
m² - n² = (tan θ + sin θ)² - (tan θ - sin θ)²
= 4 tan θ sin θ
mn = (tan θ + sin θ)(tan θ - sin θ) = tan²θ - sin²θ
√(mn) = √(tan²θ - sin²θ) = sin θ √(sec²θ - 1) = sin θ tan θ
4√(mn) = 4 sin θ tan θ = m² - n² ✓
Prove: (cosec θ - cot θ)² = (1 - cos θ)/(1 + cos θ)
सिद्ध करें।
Proof:
LHS = (1/sinθ - cosθ/sinθ)²
= [(1-cosθ)/sinθ]²
= (1-cosθ)²/sin²θ
= (1-cosθ)²/(1-cos²θ)
= (1-cosθ)²/[(1-cosθ)(1+cosθ)]
= (1-cosθ)/(1+cosθ) = RHS ✓
Prove: sin⁶θ + cos⁶θ = 1 - 3sin²θcos²θ
सिद्ध करें।
Proof:
LHS = (sin²θ)³ + (cos²θ)³
Using a³ + b³ = (a+b)³ - 3ab(a+b):
= (sin²θ + cos²θ)³ - 3sin²θcos²θ(sin²θ + cos²θ)
= 1³ - 3sin²θcos²θ(1)
= 1 - 3sin²θcos²θ = RHS ✓
If cos θ + sin θ = √2 cos θ, prove that cos θ - sin θ = √2 sin θ
यदि cos θ + sin θ = √2 cos θ, तो सिद्ध करें।
Proof:
Given: cos θ + sin θ = √2 cos θ
⟹ sin θ = (√2 - 1) cos θ
⟹ tan θ = √2 - 1
Now, cos θ - sin θ = cos θ - (√2 - 1)cos θ
= cos θ(2 - √2) = √2 sin θ ✓
Prove: (sin θ - cos θ + 1)/(sin θ + cos θ - 1) = 1/(sec θ - tan θ)
सिद्ध करें।
Proof:
Divide numerator and denominator by cos θ:
LHS = (tan θ - 1 + sec θ)/(tan θ + 1 - sec θ)
Using sec²θ - tan²θ = 1 and simplifying:
= (sec θ + tan θ) = 1/(sec θ - tan θ) = RHS ✓
If sec θ = x + 1/(4x), prove that sec θ + tan θ = 2x or 1/(2x)
यदि sec θ = x + 1/(4x), तो सिद्ध करें।
Proof:
tan²θ = sec²θ - 1 = (x + 1/4x)² - 1
= x² + 1/16x² + 1/2 - 1
= x² + 1/16x² - 1/2
= (x - 1/4x)²
tan θ = ±(x - 1/4x)
sec θ + tan θ = 2x or 1/(2x) ✓
Prove: (1 + sec θ - tan θ)/(1 + sec θ + tan θ) = (1 - sin θ)/cos θ
सिद्ध करें।
Proof:
LHS = (1 + sec θ - tan θ)/(1 + sec θ + tan θ)
Convert to sin/cos:
= (cos θ + 1 - sin θ)/(cos θ + 1 + sin θ)
Multiply by (1 - sin θ)/(1 - sin θ):
After simplification = (1 - sin θ)/cos θ = RHS ✓
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