8. त्रिकोणमिति का परिचय Mathematics Exercise - 8.3 class 10 Maths in Hindi - CBSE Study
NCERT Solutions for Class 10 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 8. त्रिकोणमिति का परिचय with detailed explanations and step-by-step answers for better exam preparation. Each प्रश्नावली 8.3 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 10 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 10 English Medium Mathematics All Chapters:
8. त्रिकोणमिति का परिचय
3. प्रश्नावली 8.3
अध्याय 8. त्रिकोणमितिय अनुपातों का परिचय
प्रश्नावली 8.3
Q1. निम्नलिखित का मान निकालिए:
(iii) cos 48° - sin 42°
हल: cos 48° - sin 42°
⇒ sin(90° - 48°) - sin 42°
⇒ sin 42° - sin 42° = 0
(iv) cosec 31° - sec 59°
हल: cosec 31° - sec 59°
⇒ sec (90° - 31°) - sec 59° [ cosec q = sec (90° - q) ]
⇒ sec 59° - sec 59° = 0
Q2. दिखाइए कि
(i) tan 48° tan 23° tan 42° tan 67° = 1
हल: (i) tan 48° tan 23° tan 42° tan 67° = 1
LHS = tan 48° tan 23° tan 42° tan 67°
= cot (90° - 48°) tan (90° - 23°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° × tan 42°) (cot 67° × tan 67°)
= 1 × 1 [ cot A × tan A = 1 ]
= 1
LHS = RHS
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
हल: (ii) cos 38° cos 52° – sin 38° sin 52° = 0
LHS = cos 38° cos 52° sin 38° sin 52°
= sin (90° - 38°) cos 52° – cos (90° - 38°) sin 52°
= sin 52° cos 52° - cos 52° sin 52°
= sin 52° (cos 52° - cos 52°)
= sin 52° × 0
= 0
LHS = RHS
Q3. यदि tan 2A = cot(A - 18°), जहाँ 2A एक न्यूनकोण है, तो A का मान ज्ञात कीजिए |
हल: tan 2A = cot(A - 18°),
⇒ cot (90° - 2A) = cot(A - 18°)
दोनों पक्षों में तुलना करने पर
⇒ 90° - 2A = A - 18°
⇒ 90° + 18° = A + 2A
⇒ 3A = 108°
Q4. यदि tan A = cot B, तो सिद्ध कीजिए कि A + B = 90°
हल: tan A = cot B दिया है |
⇒ tan A = tan (90° - B) तुलना करने पर
⇒ A = 90° - B
⇒ A + B = 90° Proved
Q5. यदि sec 4A = cosec(A - 20°), जहाँ 4A एक न्यूनकोण है, तो A का मान ज्ञात कीजिए |
हल: sec 4A = cosec(A - 20°)
⇒ cosec (90° - 4A) = cosec(A - 20°) [ sec q = (90°- q) ]
तुलना करने पर
⇒ 90° - 4A = A - 20°
⇒ 90° + 20° = A + 4A
⇒ 5A = 110°
Q7. sin 67° + cos 75° को 0° और 45° के बीच के कोणों के त्रिकोणमितिय अनुपातों के पदों में व्यक्त कीजिए |
हल : sin 67° + cos 75°
⇒ cos (90° - 67°) + sin (90° - 75°)
⇒ cos 23° + sin 15°
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